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Congruence of Triangles By AAS Criteria

Triangles I - Concepts

Calculating Perimeter of Triangle

Congruence of Triangles

Congruence of Triangles By SAS Criteria

Congruence of Triangles By ASA Criteria

Congruence of Triangles By AAS Criteria

Class - 9th IMO Subjects

Concept Explanation

Congruence of Triangles by AAS Criteria

Theorem : If any two angles and a non-included side of one triangle are equal to the corresponding angles and side of another triangle, then the two triangles are congruent.

Given: Two $large Delta sABC$ and DEF such that

$large angle A=angle D,angle B=angle E, BC=EF$

To Prove: $large Delta ABCcong Delta DEF$

Proof: We have,

$large angle A=angle D;and;angle B=angle E$

$large Rightarrow ;;angle A+angle B=angle D+angle E$

$large because angle A+angle B+angle C=180^{circ};therefore angle A+angle B=180^{circ}-angle C$

Similarly, $large because angle D+angle E+angle F=180^{circ};therefore angle D+angle E=180^{circ}-angle F$

$large Rightarrow ;;180^{circ}-angle C=180^{circ}-angle F$

Arrange the terms, we have

$angle F-angle C=180^{0}-180^{0}$

$angle F-angle C=0$

$large Rightarrow ;;;angle C=angle F$ .....................(i)

Thus, in $large Delta sABC$ and DEF , we have

$large angle A=angle D,angle B=angle E$ and $large angle C=angle F$

Now, in $large Delta ABC$ and $large Delta DEF$ , we have

$large angle B=angle E$ [Given]

BC = EF [Given]

and, $large angle C=angle F$ [From (i)]

So, by ASA criterion of congruence, $large Delta ABCcong Delta DEF$

Hence Proved

Theorem: If two angles of a triangle are equal. then sides opposite to them are also equal.

Given: A $large Delta ABC$ in which $large angle B=angle C$

To Prove: AB = AC

Construction: Draw the bisector of $large angle A$ and Let it meet BC at D.

Proof: In $large Delta sABD$ and ACD, we have

$large angle B=angle C$ [Given]

$large angle BAC=angle CAD$ [Each 90 degrees] AD = AD

So, by AAS criterion of congruence, we have

$large Delta ABDcong Delta ACD$

$large Rightarrow ;;;AB=AC$ [C.P.C.T]

Illustration: If $bigtriangleup ABC$ is an isosceles triangle with AB = AC. Prove that the perpendiculars from the vertices B and C to their opposite sides are equal.

Solution: In $bigtriangleup ABC$ , we have

AB = AC [Given]

$Rightarrow angle B=angle C$ ..........................(i) [ $because$ Angles opposite to equal sides are equal]

Now, in $bigtriangleup BCE;and;BCD$ , we have

$angle B=angle C$ [From (i)]

$angle CEB=angle BDC$ [Each equal to $90^{0}$ ]

and, BD = BC [Common]

So, by AAS criterion of congruence, we have

$bigtriangleup BCE=bigtriangleup BCD$

$Rightarrow BD=CE$ [ $because$ Corresponding parts of congruent triangles are equal]

Hence, BD = CE

Sample Questions

(More Questions for each concept available in Login)

Question : 1

In the figure above if $small angle CAB=angle CDE$ , CB = CE . If $small bigtriangleup ACBcong bigtriangleup DCE$ by AAS criterion, then by AAS criterion, then the third condition used for AAS is?
A. $small angle ACB=angle DCE$	B. $small angle B=angle E$	C. AC = CE	D. AAS is not possible
Right Option : A
View Explanation

Question : 2

In the figure given above, If $angle BAD=angle DAC;and;angle ABD=angle ACD$ . AD is common side of both these triangles then $bigtriangleup ABDcong bigtriangleup ACD$ by which criterion?
A. ASA	B. AAS	C. SAS	D. None of these
Right Option : B
View Explanation

Question : 3

In the figure shown above $angle A=angle D;,angle B=angle E;and;BC=EF$ . Then $bigtriangleup ABCcong bigtriangleup DEF$ by which criteria
A. ASA	B. AAS	C. SAS	D. AAA
Right Option : B
View Explanation