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# Congruent Triangles

Consider two given triangles PQR and ABC. They are said to be congruent (≡) if their corresponding parts are congruent. That is PQ=AB, QR=BC and PR=AC and also ∠P = ∠A, ∠Q = ∠B and ∠R = ∠C. This is denoted as ∆PQR ≡ ∆ABC .

Congruent Triangles

Consider two given triangles PQR and ABC. They are said to be congruent (≡) if their corresponding parts are congruent. That is PQ=AB, QR=BC and PR=AC and also P = A, Q = B and R = C. This is denoted as ∆PQR ≡ ∆ABC .

There are 4 ways by which one can prove that two triangles are congruent.

(i) SSS (Side – Side – Side) Congruence

If the three sides of a triangle are congruent to the three sides of another triangle, then the triangles are congruent.

That is AB = PQ, BC = QR and AC = PR

ABC PQR.

(ii) SAS (Side – Angle – Side) Congruence

If two sides and the included angle (the angle between them) of a triangle are congruent to two sides and the included angle of another triangle, then the triangles are congruent. Here, AC = PQ, A = P and AB = PR and hence ACB PQR.

(iii) ASA (Angle-Side-Angle) Congruence

If two angles and the included side of a triangle are congruent to two angles and the included side of another triangle, then the triangles are congruent. Here, A = R, CA=PR and C = P and hence ∆ABC RQP

(iv) RHS (Right Angle – Hypotenuse – Side) Congruence

If the hypotenuse and a leg of one right triangle are congruent to the hypotenuse and a leg of another right triangle, then the triangles are congruent. Here,

B = Q = 90°, BC = QR and AC = PR

(right angle) (leg) (hypotenuse)

and hence ∆ABC ≡ ∆PQR.

Note

• Any segment or angle is congruent to itself! This is called Reflexive property

• If two triangles are congruent, then their corresponding parts are congruent. This is called CPCTC (Corresponding parts of Congruent Triangles are Congruent).

• If angles then sides means if two angles are equal in a triangle, then the sides opposite to them are equal.

• If sides then angles means if two sides are equal in a triangle, then the angles opposite to them are equal.

Try these

Match the following by their congruence property

Answer: 1. (iv), 2. (iii), 3. (i), 4. (ii)

Example 5.1

Find the unknowns in the following figures

Solution:

(i) Now, from Fig. 5.7 (i), 140º + z = 180º (linear pair)

⇒ z = 180º - 140º = 40º

Also x + z = 70º  + z (exterior angle property)

x = 70º

Also z  + y + 70º = 180º (angle sum property in ΔABC)

40º + y + 70º = 180º

y = 180º - 110º = 70º

(ii) Now, from Fig. 5.7(ii), PQ = PR

Q = R (angles opposite to equal sides are equal)

x = y

x+ y + 50º = 180º (angle sum property in ΔPQR)

2 x = 130º

x = 65º

y = 65º

(iii) Now, from Fig. 5.7(iii), in ΔABC

A = x (vertically opposite angles)

Similarly B = C = x   (Why?)

A + B + C = 180º (angle sum property in ΔABC)

3 x =180º

x = 60º

y = 180º − 60º = 120º

Example 5.2 (Illustrating SSS and SAS Congruence)

If E = S and G is the midpoint of ES, prove that ∆GET GST.

Proof:

Think

In the figure, DA = DC and BA = BC.

Are the triangles DBA and DBC congruent? Why?

Solution:

AB = CB

DB = DB (common)

ΔDBA = ΔDBC        [ By SSS congruency]

Also RHS rule also bind here to say their congruency.

Example 5.3 (Illustrating ASA Congruence)

If YTB YBT and BOY TRY , prove that ∆BOY TRY

Proof:

Example 5.4 (Illustrating RHS Congruence)

If TAP is an isosceles triangle with TA = TP and TSA = 90°.

(i) Is ∆TAS TPS ? Why?

(ii) Is P = A? Why?

(iii) Is AS = PS? Why?

Proof:

(i) TA = TP (hypotenuse) and TSA = 90

TS is common (leg)

Hence, by RHS congruence, ∆TAS TPS

(ii) Given TA = TP

P = A (if angles then sides)

(iii) From (i) ∆TAS TPS ,

By CPCTC

AS=PS

SSA and ASS properties are not sucient to prove that two triangles are congruent. This is explained in the given figure. By construction, in triangles ABD and ABC, BC = BD = a. Also, AB and BAZ are common. But AC AD. So, ∆ABD is not congruent to ∆ABC and so SSA fails.

Tags : Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry
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