CBSE Class 10 Mathematics Triangles Notes Set 03

Congruent Figures and Similar Figures

Two geometric figures having the same shape and size are called congruent figures. If triangles ABC and DEF are such that AB = DE, BC = EF, AC = DF and ∠A = ∠D, ∠B = ∠E, ∠C = ∠F, then ΔABC and ΔDEF are congruent.

Figures which have the same shape but not necessarily the same size are called similar figures. For example, photographs of the same monument in stamp size, passport size and postcard size are similar. Any two circles are similar (they have the same shape but different radii), and any two squares are similar (they are congruent only when their sides are equal). A triangle and a circle are not similar, because they do not have the same shape.

From the above discussion we conclude that two congruent figures are always similar, but similar figures need not be congruent.

Similar Polygons

Two polygons having the same number of sides are said to be similar to each other if
1. their corresponding angles are equal, and
2. the lengths of their corresponding sides are proportional (in the same ratio).

If two polygons ABCDE and PQRST are similar, then ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, ∠D = ∠S, ∠E = ∠T and \( \frac{AB}{PQ}=\frac{BC}{QR}=\frac{CD}{RS}=\frac{DE}{ST}=\frac{EA}{TP} \). We write ABCDE ~ PQRST, where the symbol ~ stands for 'is similar to'.

Note: While naming similar figures, be sure to name the corresponding vertices in the same order. For example, we cannot write ABCDE ~ PQSTR, but we can write AEDCB ~ PTSRQ.

Similar Triangles

Two triangles ABC and PQR are said to be similar if
1. their corresponding angles are equal, that is ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, and
2. their corresponding sides are proportional, that is \( \frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR} \).

Note: In the case of triangles, if either of the two conditions holds, then the other holds automatically, while both conditions are needed for the similarity of other polygons.

Example 1. Check whether the following pair of triangles is similar or not: in ΔABC, ∠A = 88° and ∠B = 48°, AB = 6.7, AC = 5; in ΔPQR, ∠R = 48° and ∠Q = 42°, PQ = 13.5, PR = 10. Also check whether the rectangles with sides 12 × 6 and 9 × 4.5 are similar.

Answer:
Triangles. In ΔABC, ∠C = 180° − 88° − 48° = 44°. In ΔPQR, ∠P = 180° − 48° − 42° = 90°.
The corresponding angles are not all equal (∠A = 88° but ∠P = 90°), so the given triangles are not similar.
Rectangles. All angles are 90°. The sides are in the ratio \( \frac{6}{4.5}=\frac{4}{3}= \) and \( \frac{12}{9}=\frac{4}{3} \). The corresponding angles are equal and the sides are proportional, so the rectangles are similar.

Example 2. The parking symbol in a scale drawing is 5 inch wide and 7 inch tall. Find the width of the symbol on a sign on which it is 13 inch tall.
Answer:
The two drawings are similar, so the corresponding sides are proportional. Let the width be w inch: \( \frac{7}{5}=\frac{13}{w}\Rightarrow w=\frac{65}{7} \) inch.

Basic Proportionality Theorem (Thales Theorem)

Theorem 1. If a line is drawn parallel to one side of a triangle to intersect the other two sides at distinct points, then it divides the two sides in the same ratio.

Given: A triangle ABC in which DE ∥ BC.
To prove: \( \frac{AD}{DB}=\frac{AE}{EC} \).
Construction: Join BE and CD and draw EM ⊥ AB and DN ⊥ AC.

Proof: Since EM ⊥ AB, EM is the height of both ΔADE and ΔDBE. Hence\[ \frac{\text{ar}(ADE)}{\text{ar}(DBE)}=\frac{\tfrac12\times AD\times EM}{\tfrac12\times DB\times EM}=\frac{AD}{DB}\qquad\dots(i) \]Similarly, DN ⊥ AC is the height of ΔADE and ΔDEC, so\[ \frac{\text{ar}(ADE)}{\text{ar}(DEC)}=\frac{\tfrac12\times AE\times DN}{\tfrac12\times EC\times DN}=\frac{AE}{EC}\qquad\dots(ii) \]ΔDBE and ΔDEC are on the same base DE and between the same parallels DE and BC, so ar(DBE) = ar(DEC) ...(iii). From (i), (ii) and (iii),\[ \frac{AD}{DB}=\frac{AE}{EC} \]Hence proved.

Corollary: If in ΔABC a line DE ∥ BC intersects AB in D and AC in E, then \( \frac{AB}{AD}=\frac{AC}{AE} \) and \( \frac{AB}{DB}=\frac{AC}{EC} \).

Converse of the Basic Proportionality Theorem

Theorem 2. If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

Given: A ΔABC and a line m intersecting AB in D and AC in E such that \( \frac{AD}{DB}=\frac{AE}{EC} \).
To prove: m ∥ BC, that is DE ∥ BC.

Proof: Suppose DE is not parallel to BC. Draw DE′ ∥ BC, meeting AC at E′. By the Basic Proportionality Theorem,\[ \frac{AD}{DB}=\frac{AE'}{E'C}\qquad\dots(i) \]But it is given that\[ \frac{AD}{DB}=\frac{AE}{EC}\qquad\dots(ii) \]From (i) and (ii), \( \frac{AE'}{E'C}=\frac{AE}{EC} \). Adding 1 to both sides, \( \frac{AE'+E'C}{E'C}=\frac{AE+EC}{EC} \), that is \( \frac{AC}{E'C}=\frac{AC}{EC} \), so E′C = EC.This is possible only when E′ and E coincide, that is DE′ is the line m itself. But DE′ ∥ BC. Hence m ∥ BC. Hence proved.

Knowledge Hub
1. The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle, that is \( \frac{BD}{DC}=\frac{AB}{AC} \) (internal bisector theorem).
2. The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle, that is \( \frac{BD}{CD}=\frac{AB}{AC} \).
3. The line joining the mid-points of two sides of a triangle is parallel to the third side.
4. The diagonals of a trapezium divide each other proportionally.

Example 3. D and E are points on the sides AB and AC respectively of a ΔABC such that DE ∥ BC. Find the value of x, when AD = 4 cm, DB = (x − 4) cm, AE = 8 cm and EC = (3x − 19) cm.

Answer:
Since DE ∥ BC, by the Basic Proportionality Theorem \( \frac{AD}{DB}=\frac{AE}{EC}\Rightarrow \frac{4}{x-4}=\frac{8}{3x-19} \).
4(3x − 19) = 8(x − 4) ⇒ 12x − 76 = 8x − 32 ⇒ 4x = 44 ⇒ x = 11 cm.

Example 4. In the figure, ST ∥ QR and RS ∥ TU. Prove that PS² = PQ × PU.

Answer:
In ΔPQR, ST ∥ QR, so \( \frac{PQ}{PS}=\frac{PR}{PT} \) ...(i)
In ΔPRS, UT ∥ RS, so \( \frac{PS}{PU}=\frac{PR}{PT} \) ...(ii)
From (i) and (ii), \( \frac{PQ}{PS}=\frac{PS}{PU} \) ⇒ PS² = PQ × PU. Hence proved.

Example 5. Let P be any point on the side BC of a ΔABC. If PQ and PR are drawn parallel to BA and CA meeting CA and BA at Q and R respectively, and QR meets BC produced at T, prove that TP² = TB × TC.

Answer:
In ΔTPQ, PQ ∥ BR, so \( \frac{TB}{TP}=\frac{TR}{TQ} \) ...(i)
In ΔTQC, RP ∥ QC, so \( \frac{TP}{TC}=\frac{TR}{TQ} \) ...(ii)
From (i) and (ii), \( \frac{TB}{TP}=\frac{TP}{TC} \) ⇒ TP² = TB × TC. Hence proved.

Example 6. In the figure, ABCD is a trapezium in which AB ∥ DC. If ΔAED is similar to ΔBEC, prove that AD = BC.

Answer:
In ΔEDC and ΔEBA, ∠1 = ∠2 and ∠3 = ∠4 (alternate angles), so ΔEDC ~ ΔEBA (AA). Hence \( \frac{ED}{EB}=\frac{EC}{EA} \), that is \( \frac{ED}{EC}=\frac{EB}{EA} \) ...(i)
It is given that ΔAED ~ ΔBEC, so \( \frac{ED}{EC}=\frac{EA}{EB}=\frac{AD}{BC} \) ...(ii)
From (i) and (ii), \( \frac{EB}{EA}=\frac{EA}{EB} \) ⇒ EB² = EA² ⇒ EB = EA. Putting this in (ii), \( \frac{EA}{EA}=\frac{AD}{BC}\Rightarrow AD=BC \). Hence proved.

Example 7. In ΔABC, DE is parallel to BC meeting AB and AC at D and E respectively. If \( \frac{AD}{DB}=\frac{2}{3} \), find \( \frac{BC}{DE} \).

Answer:
Since DE ∥ BC, ∠ADE = ∠ABC and ∠A is common, so ΔABC ~ ΔADE and \( \frac{AB}{AD}=\frac{BC}{DE} \) ...(i)
From \( \frac{AD}{DB}=\frac{2}{3} \): \( \frac{DB}{AD}=\frac{3}{2}\Rightarrow \frac{DB}{AD}+1=\frac{3}{2}+1\Rightarrow \frac{AB}{AD}=\frac{5}{2} \) ...(ii)
From (i) and (ii), \( \frac{BC}{DE}=\frac{5}{2} \), that is BC : DE = 5 : 2.