Triangles — Class 10 Maths Notes

Triangles · Class 10 Maths · 26 topics.

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Topics covered in Triangles

  1. 1.Introduction of Triangles

    Let's dive into the world of triangles and understand them with simple explanations and examples.

    What is a Triangle?

    A triangle is a basic shape in geometry that has three sides and three angles. The sum of the angles in any triangle always adds up to 180 degrees.

    Types of Triangles

    1. Equilateral Triangle:

      • Definition: A triangle where all three sides are of equal length, and all three angles are equal, each being 60 degrees.
      • Example: Think of a perfectly shaped pyramid's base.

    2. Isosceles Triangle:

      • Definition: A triangle with two sides of equal length. The angles opposite these sides are also equal.
      • Example: The front face of a traditional house roof is often an isosceles triangle.

    3. Scalene Triangle:

      • Definition: A triangle where all three sides are of different lengths and all three angles are different.
      • Example: A piece of triangular cheese with no equal sides or angles.

    4. Right-Angled Triangle:

      • Definition: A triangle that has one angle measuring 90 degrees. The longest side opposite the right angle is called the 'hypotenuse'.
      • Example: The triangle formed by a ladder leaning against a wall, where the wall and the ground form a right angle.

    Real-Life Applications

    • Architecture and Construction: Understanding triangles helps in designing roofs, bridges, and structures.
    • Navigation and Astronomy: Calculating distances and positions often involves triangular calculations.
    • Art and Design: Triangles are used for creating visually appealing and stable designs.

    Activity to Understand Triangles

    1. Use sticks or straws to create different types of triangles.
    2. Measure the sides and angles to identify if it's equilateral, isosceles, scalene, or right-angled.

    Triangles are fascinating and are a fundamental part of geometry, helping us understand shapes and spaces better!

  2. 2.Similarity of Triangles

    Similarity of Triangles in Simple Terms

    Similarity in triangles is a fascinating concept in geometry. It's like comparing two pictures of the same thing, one big and one small, where the shapes are the same, but sizes are different.

    What Makes Triangles Similar?

    Two triangles are similar if they have:

    1. 1. Corresponding Angles Equal: Each angle in one triangle has the same measure as the corresponding angle in the other triangle.
    2. 2. Corresponding Sides Proportional: The ratios of the lengths of their corresponding sides are equal.

    Criteria for Triangle Similarity

    1. 1. Angle-Angle (AA) Criterion:

      • If two angles of one triangle are equal to two angles of another triangle, then the triangles are similar.

    2. 2. Side-Angle-Side (SAS) Criterion:

      • If two sides of one triangle are in the same ratio as two sides of another triangle, and the angles included between these sides are equal, the triangles are similar.

    3. 3. Side-Side-Side (SSS) Criterion:

      • If all three sides of one triangle are in the same ratio as corresponding sides of another triangle, they are similar.

    Real-Life Examples

    • Maps and Models: Similar triangles are used in making scaled maps and models. A small model of a building has the same shape as the actual building but in a smaller size.
    • Astronomy: Similar triangles help in calculating distances of stars and planets from Earth.

    Activity to Understand Triangle Similarity

    1. i. Draw two triangles of different sizes with the same angles.
    2. ii. Measure the angles and sides to see the similarity in angles and the proportionality in sides.

    Understanding similarity in triangles helps in recognizing patterns and relationships in various shapes and forms around us.

  3. 3.Theorem 1 : If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.

    To solve or prove Theorem, also known as the Basic Proportionality Theorem or Thales' Theorem, we'll walk through a step-by-step proof. This theorem is a fundamental concept in geometry and forms the basis for understanding similar triangles.

    Proof of the Basic Proportionality Theorem

    Let's consider a triangle ABC, and let's draw a line parallel to one of its sides, say side BC. Let this parallel line intersect the other two sides (AB and AC) at points D and E, respectively.

    Diagram Setup:

    • Triangle ABC with points D and E on sides AB and AC, respectively.
    • Line DE is drawn parallel to side BC.

    To Prove:

    • =DBAD​=ECAE​

    Proof Steps:

    1. Draw an Auxiliary Line: Draw a line through C and parallel to DA. Let this line intersect the extended line segment BA at point F.

    2. Using Corresponding Angles and Alternate Interior Angles: Since DE is parallel to BC, and the line through C is parallel to DA, the corresponding angles and alternate interior angles formed are equal. Therefore, ∠DAB = ∠ECF and ∠EDA = ∠EFC.

    3. Applying Similar Triangles:

      • In triangles ADE and CFE, ∠ADE = ∠CFE (by construction, both are right angles if DE is parallel to BC).
      • We have ∠DAB = ∠ECF and ∠EDA = ∠EFC (from step 2).
      • Therefore, by AA (Angle-Angle) similarity, ΔADE ∼ ΔCFE.

    4. Using Properties of Similar Triangles: In similar triangles, corresponding sides are in proportion. Thus, for ΔADE and ΔCFE:

      • =DEAD​=FECF​
      • But DE = BC (as DE is parallel and equal to BC).
      • Therefore, =BCAD​=FECF​.

    5. Applying the Triangle Proportionality Theorem: In ΔABC and the line through C parallel to DA, by the Triangle Proportionality Theorem, we have:

      • =DBAD​=CEAC​.

    6. Substituting and Concluding: Since AC = AE + EC and DE = BC,

      • =+DBAD​=ECAE+EC​.
      • Therefore, =DBAD​=ECAE​.

    Thus, the line parallel to one side of a triangle (BC in this case) divides the other two sides (AB and AC) in the same ratio. This completes the proof of Theorem 6.

    Summary

    The Basic Proportionality Theorem shows that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, those sides are divided proportionally. This theorem is a cornerstone in understanding geometric similarity and has numerous applications in different fields, including architecture, art, and engineering.

  4. 4.Theoram 2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

    Given: In ΔABC, DE is a line segment such that D is on AB, E is on AC, and DE is parallel to BC.

    To Prove: The line segment DE divides AB and AC in the same ratio, that is, =DBAD​=ECAE​.

    Proof:

    1. Since DE is parallel to BC, angle ADE is equal to angle ABC (corresponding angles with line AB).
    2. Similarly, angle AED is equal to angle ACB (corresponding angles with line AC).
    3. Now, consider triangles ADE and ABC. They have:
      • Angle ADE = angle ABC (from step 1)
      • Angle AED = angle ACB (from step 2)
      • Angle A is common to both triangles.
    4. Therefore, triangles ADE and ABC are similar by the AA (Angle-Angle) similarity criterion, which states that two triangles are similar if two angles of one triangle are equal to two angles of the other triangle.
    5. When two triangles are similar, the sides opposite the equal angles are proportional. Hence, we have: =ABAD​=ACAE​ and BCDE​.
    6. From the properties of parallel lines, DE is parallel to BC, and we can say that the sides are proportional.
    7. Rearranging the terms, we get: =DBAD​=ECAE​.

    Thus, DE divides the two sides AB and AC in the same ratio, and hence, by the converse of the Basic Proportionality Theorem, DE is parallel to BC.

    This theorem is very useful in real life, especially in fields like surveying, where similar triangles are used to measure distances indirectly. For example, if surveyors want to find the height of a mountain, they can use this theorem by creating similar triangles with the mountain and a smaller, measurable triangle.

  5. 5.If a line intersects sides AB and AC of a Triangle ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC

    Given: A triangle △△ABC with a line through D and E that is parallel to BC.

    To Prove: =ABAD​=ACAE​

    Proof:

    1. Construct the Triangle and Parallel Line: Draw △△ABC with line DE passing through D on AB and E on AC, and DE is parallel to BC.

    2. Use Similar Triangles: Because DE is parallel to BC, angle ADE is equal to angle ABC, and angle AED is equal to angle ACB (because corresponding angles with parallel lines are equal).

    3. Consider the Triangles: Look at triangles ADE and ABC. They share angle A.

      • Angle ADE is equal to angle ABC (from step 2).
      • Angle AED is equal to angle ACB (from step 2).

    4. Triangles ADE and ABC are Similar: By the Angle-Angle (AA) criterion for similarity (two pairs of corresponding angles are equal), triangle ADE is similar to triangle ABC.

    5. Write the Ratios: In similar triangles, the ratios of the corresponding sides are equal. So, for triangles ADE and ABC, we have: =ABAD​=ACAE​ because corresponding sides of similar triangles are proportional.

    6. Conclude the Proof: The ratios of the lengths of corresponding sides in similar triangles are equal, thus proving that =ABAD​=ACAE​.

    So there you have it! This theorem is quite handy in real-life scenarios like map reading, where the scale of the map creates similar triangles with real-world distances. In careers like architecture, engineering, and design, understanding how to apply the principles of similar triangles is essential for accurate model building and scaling

  6. 6.In this Figure, PS/SQ = PT/TR and ∠ PST = ∠ PRQ. Prove that PQR is an isosceles triangle.

    We are given that PS/SQ = PT/TR and that angles ∠PST and ∠PRQ are equal. We are to prove that triangle PQR is an isosceles triangle, which means we need to show that two sides of triangle PQR are equal in length.

    Here’s the proof:

    1. Given: PS/SQ = PT/TR (1) and ∠PST = ∠PRQ (2).
    2. To Prove: PQ = PR (Triangle PQR is isosceles).

    Proof:

    • From the given information, we know that lines ST and QR are parallel because corresponding angles ∠PST and ∠PRQ are equal (alternate interior angles are equal when lines are parallel).

    • With parallel lines ST and QR, and transversal lines PS and PT, we can say that triangles PST and PTR are similar because they have two angles equal (angle P is common and ∠PST = ∠PRQ).

    • In similar triangles, corresponding sides are in proportion. Therefore, from triangles PST and PTR, we have: =PTPS​=TRST​

    • But we are given that PS/SQ = PT/TR, so we can equate the two proportions because their ratios are equal. Therefore, we have: =PTPS​=TRSQ​

    • From these proportions, we can see that if we multiply the corresponding sides, we get PS * TR = PT * SQ.

    • Now, consider triangles PSQ and PTR. They are similar by the Side-Angle-Side (SAS) similarity criterion, because:

      • PS/SQ = PT/TR (given),
      • ∠PST = ∠PRQ (given, corresponding angles),
      • And we have just proven that PS * TR = PT * SQ (sides in proportion).

    • Since triangles PSQ and PTR are similar, their corresponding sides are proportional, and their corresponding angles are equal. Therefore, ∠SPQ = ∠TPR.

    • If ∠SPQ = ∠TPR, then by the converse of the Isosceles Triangle Theorem (angles opposite to equal sides are equal), PQ = PR.

    Therefore, triangle PQR must be an isosceles triangle since two of its sides, PQ and PR, are equal in length.

    This principle is quite useful in geometry, especially in the design and analysis of structures where symmetry is important, such as bridges or towers. The concept of isosceles triangles and their properties can be used to ensure stability and balance in such structures.

  7. 7.E and F are points on the sides PQ and PR respectively of a △ PQR. For each of the following cases, state whether EF || QR :

    (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

    (ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

    (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

    Solution

    To determine whether EF is parallel to QR in each case, we can use the basic proportionality theorem (also known as Thales' theorem). This theorem states that if a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides those sides proportionally.

    Let's analyze each case:

    Case i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, and FR = 2.4 cm.

    To check if EF || QR, we compare the ratios PE/EQ and PF/FR:

    =3.93=1.3EQPE​=33.9​=1.3
    =3.62.4=1.5FRPF​=2.43.6​=1.5

    Since ≠EQPE​=FRPF​, EF is not parallel to QR.

    Case ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm, and RF = 9 cm.

    Comparing the ratios PE/QE and PF/RF:

    =44.5≈0.89QEPE​=4.54​≈0.89
    =89≈0.89RFPF​=98​≈0.89

    Here, ≈QEPE​≈RFPF​, so EF is parallel to QR.

    Case iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, and PF = 0.36 cm.

    Since we have the entire lengths of PQ and PR, we need to find EQ and FR:

    EQ = PQ - PE = 1.28 - 0.18 = 1.10 cm, FR = PR - PF = 2.56 - 0.36 = 2.20 cm.

    Now, compare the ratios:

    =0.181.10EQPE​=1.100.18​ =0.362.20FRPF​=2.200.36​

    Both ratios are equal to 0.1636, so EF is parallel to QR.

    In summary, for case i, EF is not parallel to QR, while for cases ii and iii, EF is parallel to QR.

  8. 8.In Fig., if LM || CB and LN || CD, prove that AM/AB = AN/AD

    In the given figure, it is stated that LM || CB and LN || CD, and we are to prove that AM/AB = AN/AD.

    To prove this, we can use the concept of similar triangles. Here's the step-by-step explanation:

    1. Since LM || CB, by the basic proportionality theorem (also known as Thales' theorem), triangle ALM is similar to triangle ABC. This means that their corresponding sides are proportional. Therefore, we have: =(1)ABAM​=ACAL​(1)

    2. Similarly, since LN || CD, triangle ALN is similar to triangle ADC, and thus their corresponding sides are also proportional. So, we have: =(2)ADAN​=ACAL​(2)

    3. From equations (1) and (2), we can see that both ABAM​ and ADAN​ are equal to ACAL​. Therefore, it follows that: =ABAM​=ADAN​

    Hence, we have proven that AM/AB = AN/AD.

  9. 9.In Fig. DE || OQ and DF || OR. Show that EF || QR.

    The figure shows a triangle PQR with a point D on segment PQ and a point F on segment PR. Line segments DE and DF are drawn such that

    DE || OQ and DF || OR. To prove that EF is parallel to QR, we will use the basic proportionality theorem, also known as Thales' theorem.

    Thales' theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides, then it divides those sides proportionally.

    Since DE || OQ, according to Thales' theorem, we have:

    =(1)DQPD​=EOPE​(1)

    Similarly, since DF || OR, we have:

    =(2)DRPD​=FOPF​(2)

    Since both DE and DF are drawn from the same point D, PD is common in both ratios, and thus:

    =(from 1 and 2)DRDQ​=FOEO​(from 1 and 2)

    Now, if EF is parallel to QR, then by Thales' theorem again, it must divide PQ and PR proportionally, which means:

    =(3)EQPE​=FRPF​(3)

    From equation (1), we substitute /PE/EQ with /PD/DQ and from equation (2), we substitute /PF/FR with /PD/DR. So equation (3) becomes:

    =DQPD​=DRPD​

    Since /=/DQ/DR=EO/FO from the earlier step, we can say:

    =EOPD​=FOPD​

    This shows that PE and PF divide EO and FO in the same ratio, which can only be true if EF is parallel to QR, thus proving that EF || QR.

  10. 10.In Fig. A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.



    To prove that BC is parallel to QR in the given figure, where AB || PQ and AC || PR, we will once again use the basic proportionality theorem or Thales' theorem.

    Here's the step-by-step explanation:

    1. Since AB || PQ, by the basic proportionality theorem, triangle ABO is similar to triangle PQO. This means that their corresponding sides are proportional. Therefore, we have: =(1)OPAO​=PQAB​(1)

    2. Similarly, since AC || PR, triangle ACO is similar to triangle PRO. Hence, their corresponding sides are proportional as well. So, we have: =(2)OPAO​=PRAC​(2)

    3. From equations (1) and (2), since the ratios of the sides AO/OP are equal for both triangles ABO and ACO, we can infer that: =PQAB​=PRAC​

    4. This means that B and C divide the sides OQ and OR, respectively, in the same ratio.

    5. If BC is drawn to intersect OQ and OR, it must therefore do so in the same ratio according to the converse of the basic proportionality theorem.

    6. Since the line through points B and C divides the sides OQ and OR in the same ratio as AB divides PQ and AC divides PR, by the converse of the basic proportionality theorem, BC must be parallel to QR.

    Therefore, BC || QR.

  11. 11.Using Theorem 1 , prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

    To prove this statement, we can use the Converse of the Basic Proportionality Theorem (also known as Thales' theorem). The theorem states that if a line divides two sides of a triangle proportionally, then it is parallel to the third side. The converse is also true: if a line is drawn parallel to one side of a triangle, it will divide the other two sides proportionally.

    Here’s the step-by-step explanation:

    1. Let’s consider a triangle, call it ΔABC.
    2. Let D be the mid-point of side AB.
    3. We draw a line through D parallel to side BC and let it intersect side AC at point E.

    We need to prove that point E is the mid-point of side AC, meaning AE = EC.

    Since D is the mid-point of AB, we have AD = DB.

    Now, since DE is drawn parallel to BC, by the converse of the Basic Proportionality Theorem, DE will divide the other two sides proportionally. Hence, we can write that:

    =DBAD​=ECAE​

    But AD = DB because D is the mid-point of AB, so the above equation becomes:

    =ADAD​=ECAE​

    This simplifies to:

    1=1=ECAE​

    Which means AE = EC.

    Thus, E is the mid-point of AC.

    So, we have proved that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

  12. 12.Using Theorem 2 , prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side


    This statement is known as the Midpoint Theorem, or sometimes referred to as the Triangle Midsegment Theorem. The theorem states that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

    Let's prove it step by step:

    1. Consider a triangle ΔΔABC.

    2. Let D and E be the midpoints of sides AB and AC, respectively. This means =12AD=21​AB and =12AE=21​AC.

    3. We need to prove that the line segment DE is parallel to BC and that =12DE=21​BC.

    By the Converse of the Basic Proportionality Theorem (Thales' theorem), if a line divides two sides of a triangle proportionally, then it is parallel to the third side. Since D and E are midpoints, DE divides the sides AB and AC into two equal parts, thus proportionally.

    1. We draw a line through E parallel to side BC and let it intersect AB at point F.

    Since EF is parallel to BC, by the Converse of the Basic Proportionality Theorem, we have:

    =FBAF​=ECAE​

    But =AE=EC because E is the midpoint of AC, so =AF=FB, which means F is the midpoint of AB. But we already know that D is the midpoint of AB, so D and F must coincide, meaning D and F are the same point.

    1. Therefore, DE is parallel to BC because it coincides with EF, which we constructed to be parallel to BC.

    2. Additionally, since D and E are midpoints, DE must be half the length of BC because in triangle ΔΔAEC, DE is a midsegment.

    Hence, the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and half its length, proving the Midpoint Theorem.

  13. 13.ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO=CO/DO

    In trapezium ABCD, with AB || DC and the diagonals intersecting at point O, we want to prove that =BOAO​=DOCO​.

    Here's the proof using similar triangles:

    1. Draw the trapezium ABCD with AB parallel to DC and the diagonals AC and BD intersecting at point O.

    2. Because AB is parallel to DC, angle ABO is equal to angle CDO (alternate interior angles are equal when two lines are cut by a transversal).

    3. Similarly, angle BAO is equal to angle DCO for the same reason.

    4. Now, consider triangles ABO and CDO:

      • Angle ABO is equal to angle CDO (from step 2).
      • Angle BAO is equal to angle DCO (from step 3).
      • Angle O is common to both triangles. Since two angles of triangle ABO are equal to two angles of triangle CDO, the triangles are similar by the AA (Angle-Angle) postulate.

    5. When two triangles are similar, their corresponding sides are proportional. Therefore, we have:

      =COAO​=DOBO​

    6. Rearranging the terms gives us the desired relationship:

      =BOAO​=DOCO​

    This relationship is true due to the properties of similar triangles and the fact that in similar triangles, the ratios of corresponding sides are equal.

  14. 14.If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

    To prove that two triangles are similar if their corresponding angles are equal, we use the Angle-Angle (AA) Similarity Postulate. The postulate states that if two angles of one triangle are equal to two angles of another triangle, then the triangles are similar. Here's a step-by-step proof using the figure: Given: Triangle ABC and Triangle DEF are two triangles such that: ∠A = ∠D ∠B = ∠E ∠C = ∠F To Prove: Triangle ABC ∼ Triangle DEF Proof: 1. It is given that ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F. 2. By the Angle-Angle (AA) Similarity Postulate, if two angles of one triangle are congruent to two angles of another triangle, then the triangles are similar. 3. Since we have that two sets of angles are congruent (we actually have all three sets), we can say that ΔABC ∼ ΔDEF by AA postulate. Therefore, if corresponding angles are equal, the triangles are similar, which also implies that their corresponding sides are in the same ratio. Now, in the context of the diagram you provided, if line PQ is drawn parallel to side EF of Triangle DEF, such that P lies on AB and Q lies on AC, then we can say: Given: Line PQ is drawn parallel to EF, intersecting AB at P and AC at Q. To Prove: AP/PB = AQ/QC Proof: 1. Since PQ is parallel to EF and BQ intersects both, then by the Alternate Interior Angles Theorem, ∠B = ∠QPB. 2. Similarly, since PQ is parallel to EF and AP intersects both, then by the Alternate Interior Angles Theorem, ∠A = ∠PQA. 3. Now, we have two pairs of angles that are equal by construction: ∠B = ∠QPB and ∠A = ∠PQA. 4. By the AA Similarity Postulate, ΔAPQ ∼ ΔABC. 5. Therefore, the corresponding sides are in proportion: AP/AB = AQ/AC. 6. Since AB = AP + PB and AC = AQ + QC, we can find that AP/PB = AQ/QC. So, we've shown that not only are the two triangles similar, but also that the segments cut by the parallel line are proportional to the corresponding sides of the triangles. This is a part of the Triangle Proportionality Theorem. For applications, this theorem is used extensively in fields that involve measurements and scaling like cartography, architecture, and any type of design work that requires creating models or scaling objects up and down. It’s a fundamental concept in geometry that helps in understanding the properties of shapes and how they can be transformed.

  15. 15.Theoram : If in two triangles, sides of one triangle are proportional to (i.e., in the same ratio of ) the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similiar. (This criterion is referred to as

    To prove the SSS (Side-Side-Side) similarity criterion for two triangles, we'll start with two triangles where the sides of one triangle are proportional to the sides of the other triangle. We need to show that their corresponding angles are equal, which will confirm that the triangles are similar.

    Proof of the SSS Similarity Criterion:

    1. Given:

      • Triangle ABC and Triangle DEF.
      • The sides of Triangle ABC are proportional to the sides of Triangle DEF, meaning ==DEAB​=EFBC​=FDCA​.

    2. Construct:

      • Draw Triangle ABC and Triangle DEF such that AB is parallel to DE and BC is parallel to EF.

    3. Consider:

      • When lines are parallel and a transversal cuts them, the alternate interior angles are equal. So, if AB is parallel to DE and BC is parallel to EF, then angle ABC is equal to angle DEF, and angle BCA is equal to angle EFD.

    4. Use Proportionality:

      • Since the sides are proportional, we can write =DEAB​=EFBC​.

    5. Angle-Angle (AA) Similarity:

      • From steps 3 and 4, we have two pairs of equal angles. By the Angle-Angle (AA) similarity criterion, Triangle ABC is similar to Triangle DEF.

    6. Conclusion:

      • Therefore, we have shown that if the sides of two triangles are proportional, then their corresponding angles are equal, and the triangles are similar.

    This proof demonstrates the SSS similarity criterion, which is a fundamental concept in geometry.

  16. 16.If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. (This criterion is referred to as the SAS (Side–Angle–Side) similarity criterion for two tria

    To prove the SAS (Side-Angle-Side) similarity criterion for two triangles, let's consider two triangles where one angle of a triangle is equal to one angle of another triangle, and the sides including these angles are proportional. We need to prove that these two triangles are similar.

    Proof of the SAS Similarity Criterion: Given:

      • Triangle ABC and Triangle DEF.
      • Angle B of Triangle ABC is equal to Angle E of Triangle DEF (∠=∠∠B=∠E).
      • The sides around these angles are proportional, meaning =DEAB​=EFBC​.

    2. Consider:

      • We know one pair of angles is equal (∠=∠∠B=∠E).

    3. Use Proportionality:

      • The given proportionality of sides can be written as =DEAB​=EFBC​.

    4. Construct a Triangle:

      • Construct a triangle ′ABC′ such that =AB=DE and ′=BC′=EF. Since these sides are equal, △′△ABC′ is congruent to △△DEF by the SSS (Side-Side-Side) criterion of congruence.

    5. Equal Angles:

      • Since △′△ABC′ and △△DEF are congruent, their corresponding angles are equal. So, ∠=∠=∠′∠B=∠E=∠C′.

    6. Compare Triangles ABC and ABC':

      • △△ABC and △′△ABC′ have two sides in the same ratio (=BCAB​=EFDE​) and the included angle (∠∠B) equal.

    7. Use the AA (Angle-Angle) Similarity Criterion:

      • Since △△ABC and △′△ABC′ have two angles equal (∠∠B and ∠′∠C′), they are similar by the AA criterion.

    8. Conclusion:

      • Therefore, since △′△ABC′ is similar to △△ABC and △′△ABC′ is congruent to △△DEF, it follows that △△ABC is similar to △△DEF.

    This proof demonstrates that if one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, then the triangles are similar. This is the SAS similarity criterion in geometry.

  17. 17.In Given Fig. , if PQ || RS, prove that △ POQ ~ △ SOR.

    In the figure, where line PQ is parallel to line RS, we can prove that triangles POQ and SOR are similar by using the AA (Angle-Angle) criterion for similarity of triangles. This criterion states that if two angles of one triangle are equal to two angles of another triangle, the triangles are similar.

    Here's how the proof would work:

    1. Angle Correspondence:

      • Since PQ is parallel to RS and OS is a transversal, the alternate interior angles are equal. That means ∠=∠∠POQ=∠SOR by the alternate interior angle theorem.
      • Similarly, since PQ is parallel to RS and OP is a transversal, we have ∠=∠∠PQO=∠ROS by the same theorem.

    2. AA Criterion for Similarity:

      • With two pairs of corresponding angles being equal (∠=∠∠POQ=∠SOR and ∠=∠∠PQO=∠ROS), we can say by the AA similarity criterion that the triangles are similar.

    3. Conclusion:

      • Therefore, △∼△△POQ∼△SOR.

    This conclusion is based on the properties of parallel lines and the Angle-Angle criterion for the similarity of triangles.

  18. 18.In Given Fig, OA . OB = OC . OD. Show that ∠ A = ∠ C and ∠ B = ∠ D.

    In the figure, where OA multiplied by OB equals OC multiplied by OD, we can demonstrate that ∠A is equal to ∠C and ∠B is equal to ∠D by using the property of angle bisectors and the converse of the angle bisector theorem. Here's the proof explained:

    1. Given:

      • OA · OB = OC · OD
      • This implies that the products of the lengths of segments from O to the vertices of the triangle on opposite sides are equal.

    2. Angle Bisector Theorem (Converse):

      • The converse of the angle bisector theorem states that if a point in the interior of an angle has the property that the product of the distances to the sides of the angle is constant for all positions of the point, then the point lies on the bisector of the angle.

    3. Application to Triangle AOB:

      • Since OA · OB is constant, O must lie on the angle bisector of ∠A.

    4. Application to Triangle COD:

      • Since OC · OD is constant, O must lie on the angle bisector of ∠C.

    5. Equal Angles:

      • Since O lies on the angle bisector of ∠A and ∠C, and angle bisectors divide angles into two equal parts, ∠AOB is bisected into two equal angles, making ∠A = ∠B. Similarly, ∠COD is bisected, making ∠C = ∠D.

    6. Conclusion:

      • Therefore, ∠A = ∠C and ∠B = ∠D.

    This proof is based on the properties of angle bisectors and relies on the concept that the angle bisector divides an angle into two equal angles.

  19. 19.A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

    To find the length of the girl's shadow after 4 seconds, we can use the concept of similar triangles. In the figure

    To find the length of the girl's shadow after 4 seconds, we can use the concept of similar triangles. In the figure, we have two similar triangles:

    1. The large triangle formed by the lamp post and the shadow on the ground.
    2. The smaller triangle formed by the girl and her shadow.

    Let's denote:

    • AB as the height of the lamp post (3.6 m)
    • BC as the length of the girl's shadow after 4 seconds
    • DE as the height of the girl (0.9 m, since 90 cm is 0.9 m)
    • CD as the distance the girl has walked after 4 seconds

    The girl is walking away from the lamp post at a speed of 1.2 m/s, so after 4 seconds, she will have walked a distance of: =1.2×4 mCD=1.2×4 m

    The triangles ABD and CDE are similar because they are both right triangles and they share the angle at D. Therefore, the ratios of their corresponding sides are equal: =BDAB​=ECDE​

    Let's calculate CD and then use the above proportion to find BC, the length of the girl's shadow.

    First, we calculate CD: =1.2 m/s×4 sCD=1.2 m/s×4 s

    Then we substitute the known values into the proportion to solve for BC: 3.6=0.9BD3.6​=BC0.9​

    Since BD is the total distance from the base of the lamp post to the end of the shadow, BD is the sum of BC and CD.

    Let's calculate these using the information we have.

    After 4 seconds, the length of the girl's shadow will be 1.6 meters.

  20. 20.In Given Fig., CM and RN are respectively the medians of △ ABC and △ PQR. If △ ABC ~ △ PQR, prove that : (i) △AMC ~ △PNR, (ii) CM/RN = AB/PQ, (iii) △ CMB ~ △ RNQ

    Given that △△ABC is similar to △△PQR, and CM and RN are medians, we can prove the three statements as follows:

    (i) To prove △∼△△AMC∼△PNR:

    • Since CM and RN are medians, M is the midpoint of AB, and N is the midpoint of PQ.
    • In similar triangles, corresponding medians are proportional to the corresponding sides. Since △∼△△ABC∼△PQR, we have =PNAM​=PQAB​ and =PRAC​=PQAB​.
    • Therefore, two sides of △△AMC are in proportion to two corresponding sides of △△PNR, and the included angle ∠∠A is equal to ∠∠P (as △∼△
      △ABC∼△PQR).
    • By the SAS (Side-Angle-Side) similarity criterion, △∼△△AMC∼△PNR.

    (ii) To prove =RNCM​=PQAB​:

    • Since △∼△△ABC∼△PQR, the corresponding sides are in proportion, so ==PQAB​=QRBC​=PRAC​.
    • The medians of similar triangles divide the triangles into smaller triangles that are similar to the original triangle and to each other.
    • Therefore, the length of the medians will also be in the same proportion as the sides, giving us =RNCM​=PQAB​.

    (iii) To prove △∼△△CMB∼△RNQ:

    • Since △∼△△ABC∼△PQR, we have ∠=∠∠B=∠Q (as corresponding angles in similar triangles are equal).

    • M and N are midpoints of AB and PQ, respectively, so =12BM=21​AB and =12NQ=21​PQ.
    • From △∼△△ABC∼△PQR, we also have =NQBM​=PQAB​.
    • Thus, two angles are equal (∠∠B and ∠∠Q), and the sides containing these angles are in proportion.
    • By the AA (Angle-Angle) criterion, △∼△△CMB∼△RNQ.

    This completes the proof for all three parts based on the properties of similar triangles, their corresponding angles, and the proportions of their sides and medians.

  21. 21.In Fig. △ ODC ~ △ OBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

    Given that △∼△△ODC∼△OBA and the angles ∠BOC = 125° and ∠CDO = 70°, we can find the other angles as follows:

    1. Since △△ODC and △△OBA are similar, corresponding angles are equal. Thus, ∠ODC = ∠OBA.

    2. To find ∠DOC, we can use the fact that the sum of angles in a triangle is 180°. Therefore, for △△ODC: ∠ODC+∠OCD+∠CDO=180°∠ODC+∠OCD+∠CDO=180°

    Since we know ∠CDO = 70°, we can express ∠ODC and ∠OCD as follows: ∠ODC+∠OCD=180°−70°
    ∠ODC+∠OCD=180°−70° ∠ODC+∠OCD=110°
    ∠ODC+∠OCD=110°

    Now, since △△ODC is similar to △△OBA, the angle ∠ODC, which is the same as ∠OBA, must be equal to ∠DCO (since these are corresponding angles in similar triangles). Therefore, ∠ODC (or ∠OBA) and ∠OCD (or ∠DCO) are equal.

    1. In △△ODC, since ∠ODC and ∠OCD are the same and their sum is 110°, we can find each angle by dividing by 2: ∠ODC=∠OCD=110°2=55°
      ∠ODC=∠OCD=2110°​=55°

    2. To find ∠OAB, we use the fact that ∠OAB = ∠ODC due to the similarity of triangles: ∠OAB=∠ODC=55°
      ∠OAB=∠ODC=55°

    Let's summarize the angles:

    • ∠DOC is 55°.
    • ∠DCO is also 55° (since ∠ODC = ∠DCO).
    • ∠OAB is 55° (since ∠ODC = ∠OBA).

  22. 22.Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that OA/OC = OB/OD

    In the trapezium ABCD with ∥AB∥DC and diagonals AC and BD intersecting at point O, we can show that =OCOA​=ODOB​ by using similarity criteria for triangles.

    Here's how to do it:

    1. Consider triangles OAB and OCD: Since ∥AB∥DC, ∠∠OAB is equal to ∠∠OCD (alternate interior angles are equal when a transversal intersects parallel lines). Similarly, ∠∠OBA is equal to ∠∠ODC.

    2. Triangles OAB and OCD are similar: By the AA (Angle-Angle) criterion, two pairs of angles are congruent, so the triangles are similar. This can be written as △∼△△OAB∼△OCD.

    3. Proportional sides in similar triangles: In similar triangles, corresponding sides are in proportion. This means that =OCOA​=ODOB​.

    Thus, by the properties of similar triangles, we have shown that =OCOA​=ODOB​ in trapezium ABCD with diagonals intersecting at point O.

  23. 23.A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

    To find the height of the tower, we can use the concept of similar triangles. When the sun's rays are considered parallel, the triangles formed by the pole and its shadow and the tower and its shadow are similar because the corresponding angles are equal.

    Here is the step-by-step solution:

    1. Let's denote:

      • The height of the pole as ℎ=6hp​=6 m,
      • The length of the pole's shadow as =4sp​=4 m,
      • The height of the tower as ℎht​ (which we want to find),
      • The length of the tower's shadow as =28st​=28 m.

    2. Since the triangles are similar, the ratios of the heights to the lengths of their shadows will be equal. This can be expressed as: ℎ=ℎsp​hp​​=st​ht​​

    3. By substituting the known values into the equation, we get: 64=ℎ2846​=28ht​​

    4. Now, we can solve for ℎht​ by cross-multiplying: 6×28=4×ℎ6×28=4×ht​ ℎ=6×284ht​=46×28​

    Let's calculate ℎht​.

    The height of the tower is 42 meters.

  24. 24.In Fig, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ ABD ~ ∆ ECF

    To prove that △∼△△ABD∼△ECF given the conditions:

    1. △
      △ABC is isosceles with =AB=AC, which implies that ∠=∠∠ABC=∠ACB.


    2. AD is perpendicular to BC, so ∠=90∘∠ADB=90∘.


    3. EF is perpendicular to AC, so ∠=90∘∠EFC=90∘.


    4. E is a point on the extension of BC, beyond C.

    Here is the proof:

    (i) Since AD is perpendicular to BC, ∠=90∘∠ADB=90∘. Similarly, since EF is perpendicular to AC, ∠=90∘∠EFC=90∘.

    (ii) Because △△ABC is isosceles with =AB=AC, the base angles ∠∠ABC and ∠∠ACB are equal. Therefore, ∠=∠∠ABD=∠ACB.

    (iii) Since E is on the extension of BC, ∠∠ECF is an exterior angle of △△ABC at C, and by the exterior angle theorem, ∠=∠+∠∠ECF=∠ABC+∠ACB. But since △△ABC is isosceles and =AB=AC, ∠=∠∠ABC=∠ACB. Therefore, ∠∠ECF equals twice ∠∠ABC (or ∠∠ACB).

    (iv) We now know that ∠=∠=90∘∠ADB=∠EFC=90∘ and ∠=∠∠ABD=∠ECF. By the AA (Angle-Angle) similarity criterion, two pairs of angles are congruent, and thus the two triangles are similar.

    Therefore, △△ABD is similar to △△ECF by the AA similarity criterion, meaning their corresponding angles are equal and the sides around these angles are in proportion.

  25. 25.D is a point on the side BC of a triangle ABC such that ∠ADC = ∠ BAC. Show that CA² = CB.CD

    To solve this problem, we will use a geometrical approach involving similar triangles.

    Given:

    • Triangle ABC with point D on side BC.
    • ∠=∠∠ADC=∠BAC.

    We need to show:

    • 2=⋅CA2=CB⋅CD.

    Step-by-Step Solution:

    1. Angle Analysis: Since ∠=∠∠ADC=∠BAC, by the Angle-Angle (AA) criterion, triangles ACD and ABC are similar.

    2. Ratio of Sides in Similar Triangles: In similar triangles, the ratio of corresponding sides is equal. Therefore, for triangles ACD and ABC:

      =CDCA​=CACB​

    3. Rearranging the Equation: Multiplying both sides of the equation by ⋅CA⋅CD gives:

      2=⋅CA2=CB⋅CD

    This shows that 2=⋅CA2=CB⋅CD as required.

    Real-Life Application and Career Relevance: Understanding these geometric principles is fundamental in careers like architecture, engineering, and various design fields. For example, in architecture, similar concepts help in designing structures with specific proportions and angles to ensure stability and aesthetic appeal.

  26. 26.Quick Revision

    1. Introduction to Triangles:

    • A triangle is a shape with three sides and three angles.
    • The sum of its angles is always 180 degrees.
    • Types include equilateral (all sides equal), isosceles (two sides equal), and scalene (all sides different).

    2. Similarity of Triangles:

    • Triangles are similar if their corresponding angles are equal and sides are in proportion.
    • For similar triangles, the ratio of any two corresponding sides is the same.

    3. Theorem 1 (Basic Proportionality Theorem or Thales' Theorem):

    • If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.
    • Derivation: Let's take triangle ABC with a line DE parallel to BC, intersecting AB and AC. Since DE is parallel to BC, angle ADE is equal to angle ACB (alternate angles), and angle AED is equal to angle ABC. So, triangles ADE and ABC are similar, leading to AD/DB = AE/EC.

    4. Theorem 2 (Converse of Basic Proportionality Theorem):

    • If a line divides two sides of a triangle in the same ratio, then it's parallel to the third side.
    • Derivation: Take triangle ABC and a line DE such that AD/DB = AE/EC. Triangles ADE and ABC are similar (as their sides are in the same ratio). Hence, angle ADE = angle ACB and angle AED = angle ABC. Thus, DE is parallel to BC.

    Important Formulas and Implementation

    • For Similar Triangles: Side ratios are equal. For example, if ΔABC ~ ΔDEF, then AB/DE = AC/DF = BC/EF.

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