Triangles

So students, welcome to today's mathematics lesson. Today we are going to study Chapter 6 from your NCERT textbook, and that is about Triangles. Now you have been studying triangles since your early classes, and in Class IX, you learned about congruence of triangles in great detail. Let us 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 the Pythagoras Theorem that you learned earlier.

Now students, have you ever wondered how the heights of mountains, say Mount Everest, or the distances of some long distant objects like the 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. This is a very interesting application of similarity, and you will learn more about it in Chapters 8 and 9 of this book.

So let us begin by understanding what similar figures are.

In Class IX, 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. This is a very important point to remember, students.

Now 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 do you think so? Because they have different shapes altogether.

Now what can you say about 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 some photographs. You will at once say that they are the photographs of the same monument, say the Taj Mahal, but are 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.

Now what does the photographer do when she prints photographs of different sizes from the same negative? You must have heard about stamp size, passport size, and postcard size photographs. She generally takes a photograph on a small size film, say of 35 mm size, and then enlarges it into a bigger size, say 45 mm or 55 mm. 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.

So we say that two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio or proportion. This is the formal definition of similarity for polygons.

Note that the same ratio of the corresponding sides is referred to as the scale factor or the Representative Fraction for the polygons. You must have heard that world maps and blueprints for the construction of a building are prepared using a suitable scale factor and observing certain conventions.

Now students, let us perform an activity to understand similarity of figures more clearly. 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 OC, and D' lies on 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 angle A equals angle A', angle B equals angle B', angle C equals angle C', angle D equals angle D', and the ratio AB/A'B' equals BC/B'C' equals CD/C'D' equals 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.

From the above, you can easily say that quadrilaterals ABCD and PQRS are similar.

Now students, here is an important 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. This is the transitive property of similarity.

You may note that in 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, in 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. Both conditions must be satisfied simultaneously.

Now let us move on to the similarity of triangles. What can you say about the similarity of two triangles? You may recall that a 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 their corresponding angles are equal and 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, now known as the Thales Theorem, for the same.

To understand the Basic Proportionality Theorem, let us perform the following activity. Draw any angle XAY and on its one arm AX, mark points, say five points, P, Q, D, R, and B such that AP equals PQ equals QD equals DR equals 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 equals 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 triangle ABC, DE is parallel to BC and AD/DB equals AE/EC. Is it a coincidence? No, it is due to the following theorem, known as the Basic Proportionality Theorem.

Theorem 6.1 states that 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.

Now let us understand the proof of this theorem. We are given a triangle ABC in which a line parallel to side BC intersects the other two sides AB and AC at D and E respectively. We need to prove that AD/DB equals AE/EC.

Let us join BE and CD and then draw DM perpendicular to AC and EN perpendicular to AB. Now the area of triangle ADE equals half times base times height, that is half times AD times EN. Recall from Class IX that the area of triangle ADE is denoted as ar(ADE). So ar(ADE) equals half AD times EN. Similarly, ar(BDE) equals half DB times EN, ar(ADE) equals half AE times DM, and ar(DEC) equals half EC times DM.

Therefore, ar(ADE)/ar(BDE) equals (half AD times EN)/(half DB times EN) which equals AD/DB. And ar(ADE)/ar(DEC) equals (half AE times DM)/(half EC times DM) which equals AE/EC.

Note that triangles BDE and DEC are on the same base DE and between the same parallels BC and DE. So ar(BDE) equals ar(DEC). Therefore, from the three equations above, we have AD/DB equals AE/EC. And this is exactly what we wanted to prove.

Now students, is the converse of this theorem also true? Let us examine this. Draw an angle XAY on your notebook and on ray AX, mark points B1, B2, B3, B4, and B such that AB1 equals B1B2 equals B2B3 equals B3B4 equals B4B. Similarly, on ray AY, mark points C1, C2, C3, C4, and C such that AC1 equals C1C2 equals C2C3 equals C3C4 equals C4C. Then join B1C1 and BC.

Note that AB1/B1B equals AC1/C1C, each equal to 1/4. You can also see that lines B1C1 and BC are parallel to each other, that is B1C1 is parallel to BC. Similarly, by joining B2C2, B3C3, and B4C4, you can see that AB2/B2B equals AC2/C2C, which equals 2/3, and B2C2 is parallel to BC. AB3/B3B equals AC3/C3C, which equals 3/2, and B3C3 is parallel to BC. AB4/B4B equals AC4/C4C, which equals 4/1, and B4C4 is 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. You can repeat this activity by drawing any angle XAY of different measure and taking any number of equal parts on arms AX and AY. Each time, you will arrive at the same result. Thus, we obtain the following theorem, which is the converse of Theorem 6.1.

Theorem 6.2 states that 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 equals 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 equals AE'/E'C. Therefore, AE/EC equals AE'/E'C. Adding 1 to both sides of above, you can see that E and E' must coincide.

Now let us take some examples to illustrate the use of these theorems.

Example 1: 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 equals AE/AC.

Solution: DE is parallel to BC, given. So AD/DB equals AE/EC, from Theorem 6.1. Or, DB/AD equals EC/AE. Or, DB/AD plus 1 equals EC/AE plus 1. Or, AB/AD equals AC/AE. So AD/AB equals AE/AC. And this is what we wanted to prove.

Example 2: ABCD is a trapezium with AB parallel to 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 equals BF/FC.

Solution: Let us join AC to intersect EF at G. AB is parallel to DC and EF is parallel to AB, given. So EF is parallel to DC, because lines parallel to the same line are parallel to each other. Now in triangle ADC, EG is parallel to DC, as EF is parallel to DC. So AE/ED equals AG/GC, from Theorem 6.1. Similarly, from triangle CAB, CG/AG equals CF/BF, that is AG/GC equals BF/FC. Therefore, from both equations, AE/ED equals BF/FC. And this completes the proof.

Example 3: In the given figure, PS/SQ equals PT/TR and angle PST equals angle PRQ. Prove that PQR is an isosceles triangle.

Solution: It is given that PS/SQ equals PT/TR. So ST is parallel to QR, from Theorem 6.2. Therefore, angle PST equals angle PQR, because they are corresponding angles. Also, it is given that angle PST equals angle PRQ. So angle PRQ equals angle PQR. Therefore, PQ equals PR, because sides opposite the equal angles are equal. That is, PQR is an isosceles triangle.

Now students, let us move on to the criteria for similarity of triangles. In the previous section, we stated that two triangles are similar if their corresponding angles are equal and their corresponding sides are in the same ratio or proportion. That is, in triangle ABC and triangle DEF, if angle A equals angle D, angle B equals angle E, angle C equals angle F, and AB/DE equals BC/EF equals CA/FD, then the two triangles are similar.

Here, you can see that A corresponds to D, B corresponds to E, and C corresponds to F. Symbolically, we write the similarity of these two triangles as triangle ABC is similar to triangle 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 IX.

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 triangle ABC is similar to triangle EDF or triangle ABC is similar to triangle FED. However, we can write triangle BAC is similar to triangle EDF.

Now a natural question arises: for checking the similarity of two triangles, say ABC and DEF, 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 IX, you have obtained some criteria for congruency of two triangles involving only three pairs of corresponding parts or elements of the two triangles. 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 of the two triangles, instead of all the six pairs of corresponding parts.

For this, let us perform the following activity. 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 angle B equals angle E, angle C equals angle F, and angle A equals angle D. That is, corresponding angles of these two triangles are equal. What can you say about their corresponding sides? Note that BC/EF equals 3/5 equals 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, or nearly equal to 0.6 if there is some error in the measurement. Thus, AB/DE equals BC/EF equals 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 or proportion. This activity leads us to the following criterion for similarity of two triangles.

Theorem 6.3 states that 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, that is Angle-Angle-Angle, criterion of similarity of two triangles.

This theorem can be proved by taking two triangles ABC and DEF such that angle A equals angle D, angle B equals angle E, and angle C equals angle F. Cut DP equals AB and DQ equals AC and join PQ. So triangle ABC is congruent to triangle DPQ. This gives angle B equals angle P equals angle E and PQ is parallel to EF. Therefore, DP/PE equals DQ/QF. That is, AB/DE equals AC/DF. Similarly, AB/DE equals BC/EF, and so AB/DE equals BC/EF equals AC/DF.

Now here is an important 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.

Now students, 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, that is in the same ratio. 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 an activity.

Draw two triangles ABC and DEF such that AB equals 3 cm, BC equals 6 cm, CA equals 8 cm, DE equals 4.5 cm, EF equals 9 cm, and FD equals 12 cm. So you have AB/DE equals BC/EF equals CA/FD, each equal to 2/3. Now measure angle A, angle B, angle C, angle D, angle E, and angle F. You will observe that angle A equals angle D, angle B equals angle E, and angle C equals angle F, that is, the corresponding angles of the two triangles are equal. You can repeat this activity by drawing several such triangles having their sides in the same ratio. Every time you shall see that their corresponding angles are equal. It is due to the following criterion of similarity of two triangles.

Theorem 6.4 states that 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, that is Side-Side-Side, similarity criterion for two triangles.

This theorem can be proved by taking two triangles ABC and DEF such that AB/DE equals BC/EF equals CA/FD, which is less than 1. Cut DP equals AB and DQ equals AC and join PQ. It can be seen that DP/PE equals DQ/QF and PQ is parallel to EF. So angle P equals angle E and angle Q equals angle F. Therefore, DP/DE equals DQ/DF equals PQ/EF. So DP/DE equals DQ/DF equals BC/EF. So BC equals PQ. Thus, triangle ABC is congruent to triangle DPQ. So angle A equals angle D, angle B equals angle E, and angle C equals angle F.

Now here is an important 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.

Now let us recall the various criteria for congruency of two triangles learned in Class IX. 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 an activity.

Draw two triangles ABC and DEF such that AB equals 2 cm, angle A equals 50°, AC equals 4 cm, DE equals 3 cm, angle D equals 50°, and DF equals 6 cm. Here, you may observe that AB/DE equals AC/DF, each equal to 2/3, and angle A, included between the sides AB and AC, equals angle 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, that is proportion. Now let us measure angle B, angle C, angle E, and angle F. You will find that angle B equals angle E and angle C equals angle F. That is, angle A equals angle D, angle B equals angle E, and angle C equals angle F. So by AAA similarity criterion, triangle ABC is similar to triangle DEF. You may repeat this activity by drawing several pairs of such triangles with one angle of a triangle equal to one angle of another triangle and the sides including these angles are proportional. Every time, you will find that the triangles are similar. It is due to the following criterion of similarity of triangles.

Theorem 6.5 states that 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, that is 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 equals AC/DF, which is less than 1, and angle A equals angle D. Cut DP equals AB, DQ equals AC, and join PQ. Now PQ is parallel to EF and triangle ABC is congruent to triangle DPQ. So angle A equals angle D, angle B equals angle P, and angle C equals angle Q. Therefore, triangle ABC is similar to triangle DEF.

Now let us take some examples to illustrate the use of these criteria.

Example 4: In the given figure, if PQ is parallel to RS, prove that triangle POQ is similar to triangle SOR.

Solution: PQ is parallel to RS, given. So angle P equals angle S, because they are alternate angles, and angle Q equals angle R. Also, angle POQ equals angle SOR, because they are vertically opposite angles. Therefore, triangle POQ is similar to triangle SOR, by AAA similarity criterion.

Example 5: Observe the given figure and then find angle P.

Solution: In triangles ABC and PQR, AB/RQ equals 3.8/7.6 equals 1/2, BC/QP equals 6/12 equals 1/2, and CA/PR equals 3√3 divided by 6√3 equals 1/2. That is, AB/RQ equals BC/QP equals CA/PR. So triangle ABC is similar to triangle RQP, by SSS similarity. Therefore, angle C equals angle P, because they are corresponding angles of similar triangles. But angle C equals 180° minus angle A minus angle B, by the angle sum property of a triangle. That equals 180° minus 80° minus 60° equals 40°. So angle P equals 40°.

Example 6: In the given figure, OA times OB equals OC times OD. Show that angle A equals angle C and angle B equals angle D.

Solution: OA times OB equals OC times OD, given. So OA/OC equals OD/OB. Also, we have angle AOD equals angle COB, because they are vertically opposite angles. Therefore, from these two facts, triangle AOD is similar to triangle COB, by SAS similarity criterion. So angle A equals angle C and angle D equals angle B, because they are 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 equals 1.2 m multiplied by 4 equals 4.8 m. Note that in triangles ABE and CDE, angle B equals angle D, each is 90° because the lamp-post as well as the girl are standing vertical to the ground, and angle E equals angle E, the same angle. So triangle ABE is similar to triangle CDE, by AA similarity criterion. Therefore, BE/DE equals AB/CD. That is, (4.8 + x)/x equals 3.6/0.9, because 90 cm equals 90/100 m equals 0.9 m. That is, 4.8 + x equals 4x. That is, 3x equals 4.8. So x equals 1.6. So the shadow of the girl after walking for 4 seconds is 1.6 m long.

Example 8: In the given figure, CM and RN are respectively the medians of triangle ABC and triangle PQR. If triangle ABC is similar to triangle PQR, prove that triangle AMC is similar to triangle PNR, that CM/RN equals AB/PQ, and that triangle CMB is similar to triangle RNQ.

Solution: Triangle ABC is similar to triangle PQR, given. So AB/PQ equals BC/QR equals CA/RP, and angle A equals angle P, angle B equals angle Q, and angle C equals angle R. But AB equals 2 AM and PQ equals 2 PN, because CM and RN are medians. So from the first equation, 2AM/2PN equals CA/RP, that is AM/PN equals CA/RP. Also, angle MAC equals angle NPR. So from these two facts, triangle AMC is similar to triangle PNR, by SAS similarity criterion.

From this, CM/RN equals CA/RP. But CA/RP equals AB/PQ. Therefore, CM/RN equals AB/PQ.

Again, AB/PQ equals BC/QR. Therefore, CM/RN equals BC/QR. Also, CM/RN equals AB/PQ equals 2 BM/2 QN, that is CM/RN equals BM/QN. That is, CM/RN equals BC/QR equals BM/QN. Therefore, triangle CMB is similar to triangle RNQ, by SSS similarity criterion.

Now students, there is also a note to the reader about another similarity criterion. If in two right triangles, the 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 of Chapter 8, the proof will become simpler.

Now let us summarize what we have learned in this chapter.

In this chapter, you have studied the following points:

Two figures having the same shape but not necessarily the same size are called similar figures.

All the congruent figures are similar, but the converse is not true.

Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are in the same ratio, that is proportion.

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. This is the Basic Proportionality Theorem, also known as Thales Theorem.

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side. This is the converse of the Basic Proportionality Theorem.

If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio and hence the two triangles are similar. This is the AAA similarity criterion.

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. This is the AA similarity criterion, which is a consequence of AAA.

If in two triangles, corresponding sides are in the same ratio, then their corresponding angles are equal and hence the triangles are similar. This is the SSS similarity criterion.

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, that is proportional, then the triangles are similar. This is the SAS similarity criterion.

And as a note to the reader, there is also the RHS similarity criterion for right triangles, which states that if the hypotenuse and one side of one right triangle are proportional to the hypotenuse and one side of another right triangle, then the two triangles are similar.

Now students, with this we come to the end of this chapter. Make sure you understand all the similarity criteria well and practice the theorems and examples thoroughly. Thank you for your attention, and goodbye for now.