Congruence of Triangles By ASA Criteria


 
 
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

angleBAD = angleABE and angleEPA = angleDPB.Show that

(i) bigtriangleupDAP congbigtriangleupEBP

(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 angleBAD = angleEPA = angleDPB.

Prove that : (i) DeltaDAP congDeltaEBP

                      (ii) AD = BE

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

therefore In DeltaDAP and DeltaEBP,

AP = BP

Also, given angleDAP = angleEBP

and angleEPA = angleDPB

Adding angleEPD to both sides

rightarrow  angleEPA + angleEPD = angleEPD + angleDPB

rightarrow  angleAPD = angleBPE

Thus, by ASA rule

DeltaDAPcongDeltaEBP

(ii) Since, DeltaDAP cong DeltaEBP   (From above)

therefore   AD = BE     (CPCT)

Sample Questions
(More Questions for each concept available in Login)
Question : 1

In the figure given below,  small angle BCD=angle ADC and small angle ACB=angle BDA. Then small bigtriangleup ACDcong bigtriangleup BDC by which criteria.

Right Option : B
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Explanation
Question : 2

â–³ABC and â–³PQR are congruent under the correspondence ABC ↔ RQP. Write the parts of ΔABC that correspond to RP.

Right Option : A
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Explanation
Question : 3

If large DeltaABC is congruent to large DeltaDEF and ∠ B =∠ E , ∠ C =∠ F , then _________________________

Right Option : B
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Explanation
 
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