Question

in the given figure AD=AE,BD=CE prove that traingle AEB parallel to traingle ADC​

Answer 1

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