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Congruence of Triangles By ASA 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 CBSE Subjects

Concept Explanation

Congruence of Triangles by ASA Criteria

Theorem: Two triangles are congruent if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of the other triangle.

Given: $Delta ACB; and; Delta DEF$ , in which

$angle ACB; = ;angle DEF ;,; angle ABC; = ;angle DFE; and ;CB = EF$

To Prove: $Delta ACB; cong ;Delta DEF$

Proof: On comparing the sides AC and DE of $Delta ACB; and; Delta DEF$ , there are three possibilities:

(i) AC = DE (II) AC < DE (iii) AC > DE

Case (i): when AC = DE, then in $Delta ACB; and; Delta DEF$

AC = DE [ Assumed]

$angle ACB; =; angle DEF$ [ Given ]

CB = EF [Given]

Hence, $Delta ACB; cong ;Delta DEF$ [ By SAS Criterion ]

Case (ii) : When AC < DE, then we take point G on DE such that AC = GE , Join GF

Now in $Delta ACB; and; Delta GEF$

AC = GE [ Assumed]

CB = EF [ Given ]

$angle ACB; = ;angle GEF$ [Given $angle ACB; = ;angle DEF$ ]

Hence, $Delta ACB; cong ;Delta GEF$ [ By SAS Criterion ]

Hence $angle ABC; = ;angle GFE;$ [CPCT] .............(1)

But $angle ABC; = ;angle DFE;$ [ Given ] .............(2)

Therefore From (1) and (2) we get

$angle DFE; = ;angle GFE;$ But this is possible only if G and D coincide

Therefore AC = DE , Hence $Delta ACB; cong ;Delta DEF$ [ By SAS Criterion]

Case (iii): When AC > DE, then we take point G on AC such that GC = DE

Then as in Case (ii) we can prove that A coincide with G i.e. AC = DE, Hence $Delta ACB; cong ;Delta DEF$ [ By SAS Criterion]

Hence in all the three cases $Delta ACB; cong ;Delta DEF$

Illustration: AB is a line segment, AX and BY are two equal line segments drawn on opposite sides of line AB such that $large AXparallel BY$ . If AB and XY intersect each other at P , prove that

(i) $large Delta APXcong Delta BPY$ (ii) AB and XY bisect each other.

Solution: Since $large AXparallel BY$ and transversal AB intersects them at A and B respectively. Therefore,

$large angle BAX=angle ABY$ [Alternate angles]

Similarly, we have

$large angle AXY=angle BYX$ [ $large because$ Transversal XY intersects parallel lines AX and

BY at X and Y respectively]

Thus, in triangles PAX and PBY, we have

$large angle PAX=angle PBY$

$large angle AXP=angle BPY$ [Given]

and, AX = BY

So, by ASA congruence criterion, we have

$large Delta APXcong Delta BPY$

$large Rightarrow$ AP = BP and PX = PY

Hence, $large Delta APXcong Delta BPY$ and AB and XY bisect each other.

Illustration: AB is line segment and P is its mid - point. D and E are points on the same side of AB such that

$angle$ BAD = $angle$ ABE and $angle$ EPA = $angle$ DPB.Show that

(i) $bigtriangleup$ DAP $cong$ $bigtriangleup$ EBP

(ii) AD = BE

Solution:

Given: AB is a line segment and P is its mid - points. D and E are points on the same side of AB such that $angle$ BAD = $angle$ EPA = $angle$ DPB.

Prove that : (i) $Delta$ DAP $cong$ $Delta$ EBP

(ii) AD = BE

Proof: (i) Since, P is the mid - point of the line segment AB

$therefore$ In $Delta$ DAP and $Delta$ EBP,

AP = BP

Also, given $angle$ DAP = $angle$ EBP

and $angle$ EPA = $angle$ DPB

Adding $angle$ EPD to both sides

$rightarrow$ $angle$ EPA + $angle$ EPD = $angle$ EPD + $angle$ DPB

$rightarrow$ $angle$ APD = $angle$ BPE

Thus, by ASA rule

$Delta$ DAP $cong$ $Delta$ EBP

(ii) Since, $Delta$ DAP $cong$ $Delta$ EBP (From above)

$therefore$ AD = BE (CPCT)

Sample Questions

(More Questions for each concept available in Login)

Question : 1

In the below figure, if EF = QR then the congruence rule used for the congruency of the given triangles is ________________
A. SAS	B. ASA	C. RHS	D. SSS
Right Option : B
View Explanation

Question : 2

In two triangles, ABC and PQR, ∠A = 30°, ∠B = 70°, ∠P = 70°, ∠Q = 80° and AB = RP, then
A. ΔABC â‰Œ ΔPQR	B. ΔABC â‰Œ ΔQRP	C. ΔABC â‰Œ ΔRPQ	D. ΔABC â‰Œ ΔRQP
Right Option : C
View Explanation

Question : 3

In triangles ABC and PQR, if $angle A=angle R,angle B=angle P: : and : : AB=RP,$ then which one of the following congruence conditions applies :
A. SAS	B. ASA	C. SSS	D. RHS
Right Option : B
View Explanation