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:**

Here AD = CD

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 suﬃcient 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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8th Maths : Chapter 5 : Geometry : Congruent Triangles | Geometry | Chapter 5 | 8th Maths

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