Triangles

Welcome dear students! Today we are going to learn about Triangles from Class 10 Maths.

You are familiar with triangles and many of their properties from your earlier classes. In Class 9, you have studied congruence of triangles in detail. Recall that two figures are said to be congruent, if they have the same shape and the same size. In this chapter, we shall study about those figures which have the same shape but not necessarily the same size. Two figures having the same shape and not necessarily the same size are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier. Can you guess how heights of mountains, say Mount Everest, or distances of some long distant objects, say moon, have been found out? Do you think these have been measured directly with the help of a measuring tape? In fact, all these heights and distances have been found out using the idea of indirect measurements, which is based on the principle of similarity of figures.

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Now let us move on to section 6.2, Similar Figures. In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. Now consider any two or more circles. Are they congruent? Since all of them do not have the same radius, they are not congruent to each other. Note that some are congruent and some are not, but all of them have the same shape. So they all are, what we call, similar. Two similar figures have the same shape but not necessarily the same size. Therefore, all circles are similar. What about two or more squares or two or more equilateral triangles? As observed in the case of circles, here also all squares are similar and all equilateral triangles are similar. From the above, we can say that all congruent figures are similar but the similar figures need not be congruent. Can a circle and a square be similar? Can a triangle and a square be similar? These questions can be answered by just looking at the figures. Evidently these figures are not similar. Why? Because their shapes are fundamentally different.

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What can you say about the two quadrilaterals ABCD and PQRS? Are they similar? These figures appear to be similar but we cannot be certain about it. Therefore, we must have some definition of similarity of figures and based on this definition some rules to decide whether the two given figures are similar or not. For this, let us look at the photographs of the same monument, the Taj Mahal, but in different sizes. Would you say that the three photographs are similar? Yes, they are. What can you say about the two photographs of the same size of the same person, one at the age of 10 years and the other at the age of 40 years? Are these photographs similar? These photographs are of the same size but certainly they are not of the same shape. So, they are not similar. What does the photographer do when she prints photographs of different sizes from the same negative? She generally takes a photograph on a small size film, say of 35mm size and then enlarges it into a bigger size, say 45mm or 55mm. Thus, if we consider any line segment in the smaller photograph, its corresponding line segment in the bigger photograph will be 45/35 or 55/35 of that of the line segment. This really means that every line segment of the smaller photograph is enlarged in the ratio 35:45 or 35:55. It can also be said that every line segment of the bigger photograph is reduced in the ratio 45:35 or 55:35. Further, if you consider inclinations or angles between any pair of corresponding line segments in the two photographs of different sizes, you shall see that these inclinations or angles are always equal. This is the essence of the similarity of two figures and in particular of two polygons. We say that: Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio or proportion. Note that the same ratio of the corresponding sides is referred to as the scale factor or the Representative Fraction for the polygons.

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In order to understand similarity of figures more clearly, let us perform Activity 1. Place a lighted bulb at a point O on the ceiling and directly below it a table in your classroom. Let us cut a polygon, say a quadrilateral ABCD, from a plane cardboard and place this cardboard parallel to the ground between the lighted bulb and the table. Then a shadow of ABCD is cast on the table. Mark the outline of this shadow as A′B′C′D′. Note that the quadrilateral A′B′C′D′ is an enlargement or magnification of the quadrilateral ABCD. This is because of the property of light that light propagates in a straight line. You may also note that A′ lies on ray OA, B′ lies on ray OB, C′ lies on ray OC and D′ lies on ray OD. Thus, quadrilaterals A′B′C′D′ and ABCD are of the same shape but of different sizes. So, quadrilateral A′B′C′D′ is similar to quadrilateral ABCD. We can also say that quadrilateral ABCD is similar to the quadrilateral A′B′C′D′. Here, you can also note that vertex A′ corresponds to vertex A, vertex B′ corresponds to vertex B, vertex C′ corresponds to vertex C and vertex D′ corresponds to vertex D. Symbolically, these correspondences are represented as A′ ↔ A, B′ ↔ B, C′ ↔ C and D′ ↔ D. By actually measuring the angles and the sides of the two quadrilaterals, you may verify that ∠A = ∠A′, ∠B = ∠B′, ∠C = ∠C′, ∠D = ∠D′, and the ratios AB/A′B′ = BC/B′C′ = CD/C′D′ = DA/D′A′. This again emphasises that two polygons of the same number of sides are similar, if all the corresponding angles are equal and all the corresponding sides are in the same ratio or proportion.

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From the above, you can easily say that quadrilaterals ABCD and PQRS in the next figure are similar. Remark: You can verify that if one polygon is similar to another polygon and this second polygon is similar to a third polygon, then the first polygon is similar to the third polygon. You may note that in the two quadrilaterals, a square and a rectangle, corresponding angles are equal, but their corresponding sides are not in the same ratio. So, the two quadrilaterals are not similar. Similarly, you may note that in the two quadrilaterals, a square and a rhombus, corresponding sides are in the same ratio, but their corresponding angles are not equal. Again, the two polygons are not similar. Thus, either of the above two conditions of similarity of two polygons is not sufficient for them to be similar.

Now let us solve Exercise 6.1. Question 1: Fill in the blanks using the correct word given in brackets. (i) All circles are similar. (ii) All squares are similar. (iii) All equilateral triangles are similar. (iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are equal and (b) their corresponding sides are proportional. Question 2: Give two different examples of pair of (i) similar figures. (ii) non-similar figures. For similar figures, we can say any two circles, or any two squares. For non-similar figures, we can say a circle and a triangle, or a square and a rectangle. Question 3: State whether the following quadrilaterals are similar or not. The figure shows a square and a rectangle. Their corresponding angles are all 90°, but their corresponding sides are not in the same ratio. Therefore, they are not similar.

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Now we move to section 6.3, Similarity of Triangles. What can you say about the similarity of two triangles? You may recall that triangle is also a polygon. So, we can state the same conditions for the similarity of two triangles. That is: Two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio or proportion. Note that if corresponding angles of two triangles are equal, then they are known as equiangular triangles. A famous Greek mathematician Thales gave an important truth relating to two equiangular triangles which is as follows: The ratio of any two corresponding sides in two equiangular triangles is always the same. It is believed that he had used a result called the Basic Proportionality Theorem for the same. To understand the Basic Proportionality Theorem, let us perform Activity 2. Draw any angle XAY and on its one arm AX, mark points, say five points, P, Q, D, R and B such that AP = PQ = QD = DR = RB. Now, through B, draw any line intersecting arm AY at C. Also, through the point D, draw a line parallel to BC to intersect AC at E. Do you observe from your constructions that AD/DB = 3/2? Measure AE and EC. What about AE/EC? Observe that AE/EC is also equal to 3/2. Thus, you can see that in ∆ABC, DE || BC and AD/DB = AE/EC. Is it a coincidence? No, it is due to the following theorem, known as the Basic Proportionality Theorem.

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Theorem 6.1: 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. Proof: We are given a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively. We need to prove that AD/DB = AE/EC. Let us join BE and CD and then draw DM ⊥ AC and EN ⊥ AB. Now, area of ∆ADE = 1/2 base × height, which is 1/2 AD × EN. Recall from Class 9, that area of ∆ADE is denoted as ar(ADE). So, ar(ADE) = 1/2 AD × EN. Similarly, ar(BDE) = 1/2 DB × EN, ar(ADE) = 1/2 AE × DM and ar(DEC) = 1/2 EC × DM. Therefore, ar(ADE)/ar(BDE) = (1/2 AD × EN) / (1/2 DB × EN), which simplifies to AD/DB. And ar(ADE)/ar(DEC) = (1/2 AE × DM) / (1/2 EC × DM), which simplifies to AE/EC. Note that ∆BDE and ∆DEC are on the same base DE and between the same parallels BC and DE. So, ar(BDE) = ar(DEC). Therefore, from the first two equations and this third fact, we have AD/DB = AE/EC.

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Is the converse of this theorem also true? To examine this, let us perform Activity 3. Draw an angle XAY on your notebook and on ray AX, mark points B₁, B₂, B₃, B₄ and B such that AB₁ = B₁B₂ = B₂B₃ = B₃B₄ = B₄B. Similarly, on ray AY, mark points C₁, C₂, C₃, C₄ and C such that AC₁ = C₁C₂ = C₂C₃ = C₃C₄ = C₄C. Then join B₁C₁ and BC. Note that AB₁/B₁B = AC₁/C₁C, each equal to 1/4. You can also see that lines B₁C₁ and BC are parallel to each other. Similarly, by joining B₂C₂, B₃C₃ and B₄C₄, you can see that the ratios are 2/3, 3/2, and 4/1 respectively, and in each case the lines are parallel to BC. From these observations, it can be observed that if a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. Thus, we obtain the following theorem, which is the converse of Theorem 6.1.

Theorem 6.2: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. This theorem can be proved by taking a line DE such that AD/DB = AE/EC and assuming that DE is not parallel to BC. If DE is not parallel to BC, draw a line DE′ parallel to BC. So, AD/DB = AE′/E′C. Therefore, AE/EC = AE′/E′C. Adding 1 to both sides of above, you can see that E and E′ must coincide. Hence, DE is parallel to BC.

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Let us take some examples to illustrate the use of the above theorems. Example 1: If a line intersects sides AB and AC of a ∆ABC at D and E respectively and is parallel to BC, prove that AD/AB = AE/AC. Solution: DE || BC, given. So, AD/DB = AE/EC by Theorem 6.1. Taking reciprocals, DB/AD = EC/AE. Adding 1 to both sides, DB/AD + 1 = EC/AE + 1. This gives (DB+AD)/AD = (EC+AE)/AE. So, AB/AD = AC/AE. Therefore, AD/AB = AE/AC.

Example 2: ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Show that AE/ED = BF/FC. Solution: Let us join AC to intersect EF at G. AB || DC and EF || AB, given. So, EF || DC, because lines parallel to the same line are parallel to each other. Now, in ∆ADC, EG || DC. So, AE/ED = AG/GC by Theorem 6.1. Similarly, from ∆CAB, CG/AG = CF/BF, which means AG/GC = BF/FC. Therefore, from these two equations, AE/ED = BF/FC.

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Example 3: In the given figure, PS/SQ = PT/TR and ∠PST = ∠PRQ. Prove that PQR is an isosceles triangle. Solution: It is given that PS/SQ = PT/TR. So, ST || QR by Theorem 6.2. Therefore, ∠PST = ∠PQR, as they are corresponding angles. Also, it is given that ∠PST = ∠PRQ. So, ∠PRQ = ∠PQR. Therefore, PQ = PR, because sides opposite the equal angles are equal. Hence, PQR is an isosceles triangle.

Now let us solve Exercise 6.2. Question 1: In the figures, DE || BC. Find EC in the first figure and AD in the second. In the first figure, AD = 1.5, DB = 3, and AE = 1. By Theorem 6.1, AD/DB = AE/EC. So, 1.5/3 = 1/EC. This gives 0.5 = 1/EC, so EC = 2. In the second figure, AD is unknown, DB = 7.2, AE = 1.8, and EC = 5.4. AD/7.2 = 1.8/5.4. 1.8/5.4 is 1/3. So AD = 7.2/3, which is 2.4.

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Question 2: E and F are points on the sides PQ and PR respectively of a ∆PQR. For each case, state whether EF || QR. Part (i): PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm. PE/EQ = 3.9/3 = 1.3. PF/FR = 3.6/2.4 = 1.5. Since the ratios are not equal, EF is not parallel to QR. Part (ii): PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm. PE/EQ = 4/4.5 = 8/9. PF/FR = 8/9. Since they are equal, EF || QR. Part (iii): PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm. EQ = PQ - PE = 1.28 - 0.18 = 1.1. FR = PR - PF = 2.56 - 0.36 = 2.2. PE/EQ = 0.18/1.1 = 18/110 = 9/55. PF/FR = 0.36/2.2 = 36/220 = 9/55. Since they are equal, EF || QR.

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Question 3: In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD. Solution: In ∆ABC, LM || CB. By Theorem 6.1, AM/MB = AL/LC. In ∆ADC, LN || CD. By Theorem 6.1, AN/ND = AL/LC. Therefore, AM/MB = AN/ND. Adding 1 to both sides, AM/MB + 1 = AN/ND + 1. This gives (AM+MB)/MB = (AN+ND)/ND. So, AB/MB = AD/ND. Taking reciprocals, MB/AB = ND/AD. Subtracting from 1, 1 - MB/AB = 1 - ND/AD. This gives (AB-MB)/AB = (AD-ND)/AD. So, AM/AB = AN/AD.

Question 4: In the figure, DE || AC and DF || AE. Prove that BF/FE = BE/EC. Solution: In ∆ABC, DE || AC. So, BD/DA = BE/EC. In ∆ABE, DF || AE. So, BD/DA = BF/FE. Therefore, BE/EC = BF/FE.

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Question 5: In the figure, DE || OQ and DF || OR. Show that EF || QR. Solution: In ∆POQ, DE || OQ. So, PD/DO = PE/EQ. In ∆POR, DF || OR. So, PD/DO = PF/FR. Therefore, PE/EQ = PF/FR. By the converse of Basic Proportionality Theorem, EF || QR.

Question 6: In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Solution: In ∆OPQ, AB || PQ. So, OA/AP = OB/BQ. In ∆OPR, AC || PR. So, OA/AP = OC/CR. Therefore, OB/BQ = OC/CR. By the converse of Basic Proportionality Theorem, BC || QR.

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Question 7: Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. Solution: Let D be the mid-point of side AB of ∆ABC. So, AD = DB. Draw a line through D parallel to BC, intersecting AC at E. By Theorem 6.1, AD/DB = AE/EC. Since AD = DB, AD/DB = 1. So, AE/EC = 1, which means AE = EC. Hence, E is the mid-point of AC. The line bisects the third side.

Question 8: Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. Solution: Let D and E be the mid-points of sides AB and AC of ∆ABC. So, AD = DB and AE = EC. This means AD/DB = 1 and AE/EC = 1. So, AD/DB = AE/EC. By Theorem 6.2, DE || BC.

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Question 9: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/CO = BO/DO. Solution: In ∆OAB and ∆OCD, AB || DC. So, ∠OAB = ∠OCD and ∠OBA = ∠ODC as alternate interior angles. Also, ∠AOB = ∠COD as vertically opposite angles. So, ∆OAB ~ ∆OCD by AAA criterion. Therefore, corresponding sides are proportional: AO/CO = BO/DO.

Question 10: The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/CO = BO/DO. Show that ABCD is a trapezium. Solution: Given AO/CO = BO/DO. In ∆AOB and ∆COD, we have AO/CO = BO/DO and ∠AOB = ∠COD vertically opposite. By SAS similarity, ∆AOB ~ ∆COD. Therefore, ∠OAB = ∠OCD. These are alternate interior angles for lines AB and DC with transversal AC. Hence, AB || DC. Since one pair of opposite sides is parallel, ABCD is a trapezium.

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Now we move to section 6.4, Criteria for Similarity of Triangles. In the previous section, we stated that two triangles are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio. That is, in ∆ABC and ∆DEF, if ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, and AB/DE = BC/EF = CA/FD, then the two triangles are similar. Here, A corresponds to D, B corresponds to E and C corresponds to F. Symbolically, we write the similarity of these two triangles as ∆ABC ~ ∆DEF and read it as triangle ABC is similar to triangle DEF. The symbol ~ stands for is similar to. Recall that you have used the symbol ≅ for is congruent to in Class 9. It must be noted that as done in the case of congruency of two triangles, the similarity of two triangles should also be expressed symbolically, using correct correspondence of their vertices. For example, for the triangles ABC and DEF, we cannot write ∆ABC ~ ∆EDF or ∆ABC ~ ∆FED. However, we can write ∆BAC ~ ∆EDF.

Now a natural question arises: For checking the similarity of two triangles, should we always look for all the equality relations of their corresponding angles and all the equality relations of the ratios of their corresponding sides? Let us examine. You may recall that in Class 9, you have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts. Here also, let us make an attempt to arrive at certain criteria for similarity of two triangles involving relationship between less number of pairs of corresponding parts. For this, let us perform Activity 4. Draw two line segments BC and EF of two different lengths, say 3 cm and 5 cm respectively. Then, at the points B and C respectively, construct angles PBC and QCB of some measures, say 60° and 40°. Also, at the points E and F, construct angles REF and SFE of 60° and 40° respectively. Let rays BP and CQ intersect each other at A and rays ER and FS intersect each other at D. In the two triangles ABC and DEF, you can see that ∠B = ∠E, ∠C = ∠F and ∠A = ∠D. That is, corresponding angles of these two triangles are equal. What can you say about their corresponding sides? Note that BC/EF = 3/5 = 0.6. What about AB/DE and CA/FD? On measuring AB, DE, CA and FD, you will find that AB/DE and CA/FD are also equal to 0.6. Thus, AB/DE = BC/EF = CA/FD. You can repeat this activity by constructing several pairs of triangles having their corresponding angles equal. Every time, you will find that their corresponding sides are in the same ratio. This activity leads us to the following criterion for similarity of two triangles.

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Theorem 6.3: 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. This criterion is referred to as the AAA (Angle–Angle–Angle) criterion of similarity of two triangles. This theorem can be proved by taking two triangles ABC and DEF such that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. Cut DP = AB and DQ = AC and join PQ. So, ∆ABC ≅ ∆DPQ by SAS congruence. This gives ∠B = ∠P = ∠E and PQ || EF. Therefore, DP/PE = DQ/QF, which means AB/DE = AC/DF. Similarly, AB/DE = BC/EF and so AB/DE = BC/EF = AC/DF.

Remark: If two angles of a triangle are respectively equal to two angles of another triangle, then by the angle sum property of a triangle their third angles will also be equal. Therefore, AAA similarity criterion can also be stated as follows: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. This may be referred to as the AA similarity criterion for two triangles.

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You have seen above that if the three angles of one triangle are respectively equal to the three angles of another triangle, then their corresponding sides are proportional. What about the converse of this statement? Is the converse true? In other words, if the sides of a triangle are respectively proportional to the sides of another triangle, is it true that their corresponding angles are equal? Let us examine it through Activity 5. Draw two triangles ABC and DEF such that AB = 3 cm, BC = 6 cm, CA = 8 cm, DE = 4.5 cm, EF = 9 cm and FD = 12 cm. So, you have AB/DE = BC/EF = CA/FD, each equal to 2/3. Now measure ∠A, ∠B, ∠C, ∠D, ∠E and ∠F. You will observe that ∠A = ∠D, ∠B = ∠E and ∠C = ∠F. It is due to the following criterion of similarity of two triangles: Theorem 6.4: If in two triangles, sides of one triangle are proportional to, that is, in the same ratio of, the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This criterion is referred to as the SSS (Side–Side–Side) similarity criterion for two triangles. This theorem can be proved by taking two triangles ABC and DEF such that AB/DE = BC/EF = CA/FD < 1. Cut DP = AB and DQ = AC and join PQ. It can be seen that DP/PE = DQ/QF and PQ || EF. So, ∠P = ∠E and ∠Q = ∠F. Therefore, DP/DE = DQ/DF = PQ/EF. So, DP/DE = DQ/DF = BC/EF. So, BC = PQ. Thus, ∆ABC ≅ ∆DPQ. So, ∠A = ∠D, ∠B = ∠E and ∠C = ∠F.

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Remark: You may recall that either of the two conditions namely, corresponding angles are equal and corresponding sides are in the same ratio is not sufficient for two polygons to be similar. However, on the basis of Theorems 6.3 and 6.4, you can now say that in case of similarity of the two triangles, it is not necessary to check both the conditions as one condition implies the other. Let us now recall the various criteria for congruency of two triangles learnt in Class 9. You may observe that SSS similarity criterion can be compared with the SSS congruency criterion. This suggests us to look for a similarity criterion comparable to SAS congruency criterion of triangles. For this, let us perform Activity 6. Draw two triangles ABC and DEF such that AB = 2 cm, ∠A = 50°, AC = 4 cm, DE = 3 cm, ∠D = 50° and DF = 6 cm. Here, you may observe that AB/DE = AC/DF, each equal to 2/3, and ∠A, included between the sides AB and AC, equals ∠D, included between the sides DE and DF. That is, one angle of a triangle is equal to one angle of another triangle and sides including these angles are in the same ratio. Now let us measure ∠B, ∠C, ∠E and ∠F. You will find that ∠B = ∠E and ∠C = ∠F. So, by AAA similarity criterion, ∆ABC ~ ∆DEF. You may repeat this activity by drawing several pairs of such triangles. Everytime, you will find that the triangles are similar. It is due to the following criterion of similarity of triangles: Theorem 6.5: 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 triangles. As before, this theorem can be proved by taking two triangles ABC and DEF such that AB/DE = AC/DF < 1 and ∠A = ∠D. Cut DP = AB, DQ = AC and join PQ. Now, PQ || EF and ∆ABC ≅ ∆DPQ. So, ∠A = ∠D, ∠B = ∠P and ∠C = ∠Q. Therefore, ∆ABC ~ ∆DEF.

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We now take some examples to illustrate the use of these criteria. Example 4: In the figure, if PQ || RS, prove that ∆POQ ~ ∆SOR. Solution: PQ || RS, given. So, ∠P = ∠S as alternate angles and ∠Q = ∠R. Also, ∠POQ = ∠SOR as vertically opposite angles. Therefore, ∆POQ ~ ∆SOR by AAA similarity criterion.

Example 5: Observe the figure and then find ∠P. Solution: In ∆ABC and ∆PQR, AB/RQ = 3.8/7.6 = 1/2. BC/QP = 6/12 = 1/2. And CA/PR = 3/6 = 1/2. That is, AB/RQ = BC/QP = CA/PR. So, ∆ABC ~ ∆RQP by SSS similarity. Therefore, ∠C = ∠P as corresponding angles of similar triangles. But ∠C = 180° - ∠A - ∠B by angle sum property. So, ∠C = 180° - 80° - 60° = 40°. Therefore, ∠P = 40°.

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Example 6: In the figure, OA × OB = OC × OD. Show that ∠A = ∠C and ∠B = ∠D. Solution: OA × OB = OC × OD, given. So, OA/OC = OD/OB. Also, we have ∠AOD = ∠COB as vertically opposite angles. Therefore, from these two, ∆AOD ~ ∆COB by SAS similarity criterion. So, ∠A = ∠C and ∠D = ∠B as corresponding angles of similar triangles.

Example 7: 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. Solution: Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post. From the figure, you can see that DE is the shadow of the girl. Let DE be x metres. Now, BD = 1.2 m × 4 = 4.8 m. Note that in ∆ABE and ∆CDE, ∠B = ∠D, each is 90° because lamp-post as well as the girl are standing vertical to the ground, and ∠E = ∠E as same angle. So, ∆ABE ~ ∆CDE by AA similarity criterion. Therefore, BE/DE = AB/CD. That is, (4.8 + x)/x = 3.6/0.9. Note that 90 cm = 0.9 m. So, (4.8 + x)/x = 4. This gives 4.8 + x = 4x. So, 3x = 4.8. Hence, x = 1.6. So, the shadow of the girl after walking for 4 seconds is 1.6 m long.

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Example 8: In the figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that (i) ∆AMC ~ ∆PNR, (ii) CM/RN = AB/PQ, and (iii) ∆CMB ~ ∆RNQ. Solution: Part (i): ∆ABC ~ ∆PQR, given. So, AB/PQ = BC/QR = CA/RP, and ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R. But AB = 2AM and PQ = 2PN, as CM and RN are medians. So, from the first ratio, 2AM/2PN = CA/RP, which gives AM/PN = CA/RP. Also, ∠MAC = ∠NPR. So, from these, ∆AMC ~ ∆PNR by SAS similarity. Part (ii): From part (i), CM/RN = CA/RP. But CA/RP = AB/PQ. Therefore, CM/RN = AB/PQ. Part (iii): Again, AB/PQ = BC/QR. Therefore, CM/RN = BC/QR. Also, CM/RN = 2BM/2QN = BM/QN. So, CM/RN = BC/QR = BM/QN. Therefore, ∆CMB ~ ∆RNQ by SSS similarity.

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Now let us solve Exercise 6.3. Question 1: State which pairs of triangles in the figure are similar. Write the similarity criterion used and write the pairs in symbolic form. In the first pair, angles are 70°, 50°, and 60° in both. So, they are similar by AAA criterion. Symbolically, ∆ABC ~ ∆PQR. In the second pair, sides are proportional: 2/2.5 = 3/3.75 = 4/5, all equal to 0.8. So, similar by SSS criterion. Symbolically, ∆ABC ~ ∆DEF. In the third pair, two sides are proportional but the included angle is not equal, so not similar. In the fourth pair, two angles are equal, so similar by AA criterion. Symbolically, ∆MNO ~ ∆XYZ.

Question 2: In the figure, ∆ODC ~ ∆OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB. Solution: ∠DOC and ∠BOC form a linear pair, so ∠DOC = 180° - 125° = 55°. In ∆ODC, ∠DCO = 180° - ∠CDO - ∠DOC = 180° - 70° - 55° = 55°. Since ∆ODC ~ ∆OBA, ∠OAB = ∠OCD = 55°.

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Question 3: 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. Solution: In ∆OAB and ∆OCD, AB || DC. So, ∠OAB = ∠OCD and ∠OBA = ∠ODC as alternate interior angles. Therefore, ∆OAB ~ ∆OCD by AA similarity. Hence, OA/OC = OB/OD.

Question 4: In the figure, QR/QT = QS/PR and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR. Solution: Given ∠1 = ∠2, so PQ = PR. Also, QR/QT = QS/PR, which means QR/QT = QS/PQ. In ∆PQS and ∆TQR, ∠Q is common. And the sides including ∠Q are proportional: QS/QR = PQ/QT. So, ∆PQS ~ ∆TQR by SAS similarity.

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Question 5: S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS. Solution: In ∆RPQ and ∆RTS, ∠P = ∠RTS, given. ∠R is common to both triangles. Therefore, ∆RPQ ~ ∆RTS by AA similarity criterion.

Question 6: In the figure, if ∆ABE ≅ ∆ACD, show that ∆ADE ~ ∆ABC. Solution: Since ∆ABE ≅ ∆ACD, AB = AC and AE = AD. So, AB/AC = 1 and AE/AD = 1. Thus, AB/AC = AE/AD. Also, ∠A is common. Therefore, ∆ADE ~ ∆ABC by SAS similarity.

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Question 7: In the figure, altitudes AD and CE of ∆ABC intersect each other at the point P. Show that (i) ∆AEP ~ ∆CDP, (ii) ∆ABD ~ ∆CBE, (iii) ∆AEP ~ ∆ADB, (iv) ∆PDC ~ ∆BEC. Solution: Part (i): In ∆AEP and ∆CDP, ∠AEP = ∠CDP = 90°. ∠APE = ∠CPD as vertically opposite. So, similar by AA. Part (ii): In ∆ABD and ∆CBE, ∠ADB = ∠CEB = 90°. ∠B is common. So, similar by AA. Part (iii): In ∆AEP and ∆ADB, ∠AEP = ∠ADB = 90°. ∠A is common. So, similar by AA. Part (iv): In ∆PDC and ∆BEC, ∠PDC = ∠BEC = 90°. ∠C is common. So, similar by AA.

Question 8: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ∆ABE ~ ∆CFB. Solution: In parallelogram ABCD, AB || DC and AD || BC. Since E is on AD produced, AE || BC. In ∆ABE and ∆CFB, ∠ABE = ∠CFB as alternate interior angles. ∠BAE = ∠FCB as opposite angles of parallelogram are equal. So, ∆ABE ~ ∆CFB by AA similarity.

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Question 9: In the figure, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that (i) ∆ABC ~ ∆AMP, (ii) CA/BC = PA/MP. Solution: Part (i): In ∆ABC and ∆AMP, ∠ABC = ∠AMP = 90°. ∠A is common. So, ∆ABC ~ ∆AMP by AA similarity. Part (ii): Since the triangles are similar, corresponding sides are proportional. So, CA/PA = BC/MP. Rearranging, CA/BC = PA/MP.

Question 10: CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively. If ∆ABC ~ ∆FEG, show that (i) CD/AC = GH/FG, (ii) ∆DCB ~ ∆HGE, (iii) ∆DCA ~ ∆HGF. Solution: Part (i): Since ∆ABC ~ ∆FEG, ∠A = ∠F, ∠B = ∠E, and ∠C = ∠G. CD and GH are angle bisectors, so ∠ACD = 1/2 ∠C and ∠FGH = 1/2 ∠G. Since ∠C = ∠G, ∠ACD = ∠FGH. In ∆ACD and ∆FGH, ∠A = ∠F and ∠ACD = ∠FGH. So, they are similar by AA. Hence, CD/GH = AC/FG, which rearranges to CD/AC = GH/FG. Part (ii): In ∆DCB and ∆HGE, ∠B = ∠E and ∠DCB = 1/2 ∠C = 1/2 ∠G = ∠HGE. So, similar by AA. Part (iii): In ∆DCA and ∆HGF, ∠A = ∠F and ∠DCA = ∠HGF. So, similar by AA.

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Question 11: In the figure, 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. Solution: In isosceles triangle ABC, AB = AC, so ∠B = ∠C. AD ⊥ BC, so ∠ADB = 90°. EF ⊥ AC, so ∠EFC = 90°. In ∆ABD and ∆ECF, ∠B = ∠C, and ∠ADB = ∠EFC = 90°. Therefore, ∆ABD ~ ∆ECF by AA similarity.

Question 12: Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. Show that ∆ABC ~ ∆PQR. Solution: Given AB/PQ = BC/QR = AD/PM. Since AD and PM are medians, BD = 1/2 BC and QM = 1/2 QR. So, AB/PQ = BD/QM = AD/PM. Thus, ∆ABD ~ ∆PQM by SSS similarity. So, ∠B = ∠Q. Now in ∆ABC and ∆PQR, AB/PQ = BC/QR and ∠B = ∠Q. Therefore, ∆ABC ~ ∆PQR by SAS similarity.

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Question 13: D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB × CD. Solution: In ∆ADC and ∆BAC, ∠ADC = ∠BAC, given. ∠C is common. So, ∆ADC ~ ∆BAC by AA similarity. Therefore, corresponding sides are proportional: CA/CB = CD/CA. Cross multiplying, CA² = CB × CD.

Question 14: Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ∆ABC ~ ∆PQR. Solution: Given AB/PQ = AC/PR = AD/PM. Extend AD to E such that AD = DE. Join BE and CE. Similarly, extend PM to N such that PM = MN. Join QN and RN. By parallelogram properties, AB = CE and AC = BE. Similarly, PQ = RN and PR = QN. So, CE/RN = BE/QN = AE/PN. Thus, ∆ABE ~ ∆PQN by SSS. So, ∠BAE = ∠QPN. Similarly, ∆AEC ~ ∆PNR, so ∠CAE = ∠RPM. Adding, ∠BAC = ∠QPR. Now in ∆ABC and ∆PQR, AB/PQ = AC/PR and ∠BAC = ∠QPR. So, ∆ABC ~ ∆PQR by SAS similarity.

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Question 15: 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. Solution: Let the height of the tower be h metres. The pole and tower are vertical, so they form right angles with the ground. The sun rays are parallel, so the angles of elevation are equal. Thus, the triangle formed by the pole and its shadow is similar to the triangle formed by the tower and its shadow by AA similarity. So, height of pole / height of tower = shadow of pole / shadow of tower. That is, 6/h = 4/28. Cross multiplying, 4h = 168. So, h = 42. The height of the tower is 42 metres.

Question 16: If AD and PM are medians of triangles ABC and PQR, respectively where ∆ABC ~ ∆PQR, prove that AB/PQ = AD/PM. Solution: Since ∆ABC ~ ∆PQR, AB/PQ = BC/QR and ∠B = ∠Q. Since AD and PM are medians, BD = 1/2 BC and QM = 1/2 QR. So, AB/PQ = BD/QM. In ∆ABD and ∆PQM, AB/PQ = BD/QM and ∠B = ∠Q. So, ∆ABD ~ ∆PQM by SAS similarity. Therefore, AB/PQ = AD/PM.

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Now let us review the summary of this chapter. In this chapter you have studied the following points. 1. Two figures having the same shape but not necessarily the same size are called similar figures. 2. All the congruent figures are similar but the converse is not true. 3. Two polygons of the same number of sides are similar, if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio. 4. If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio. 5. If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. 6. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar (AAA similarity criterion). 7. If in two triangles, two angles of one triangle are respectively equal to the two angles of the other triangle, then the two triangles are similar (AA similarity criterion). 8. If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar (SSS similarity criterion). 9. If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are in the same ratio, then the triangles are similar (SAS similarity criterion).

A note to the reader: If in two right triangles, hypotenuse and one side of one triangle are proportional to the hypotenuse and one side of the other triangle, then the two triangles are similar. This may be referred to as the RHS Similarity Criterion. If you use this criterion in Example 2, Chapter 8, the proof will become simpler.

Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]