Question

In adjacent fig , AE=AD & BD=CE . PROVE THAT TRIANGLE AEB CONGURENCE TRIANGLE ADC.

Answer 1

In    ΔAEB    and    ΔADC
     
          AE =  AD          (Given)
     ∠EAB = ∠DAC      (common)
         AB  =  AC          { :- AE = AD and EC = BD }

thus

Δ AEB ≈  ΔADC  (SAS congurency)

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Answer 2

Solutions:

We have,

AE = AD and CE = BD

=> AE + CE = AD + BD

=> AC = AB ............... (i)

Thus, in △'sAEB and ADC, we have

AE = AD ............ [Given]

∠EAB = ∠DAC .......[Common]

and, AC = AB ......... (i)

So, by SAS criterion of congruence, we obtain

△AEB ≅△ADC