Question

In triangle CAB, D and E are points on CA and CB respectively such that angle A = angle CED.

Prove that:-
triangle CAB is similar to triangle CED.


Please answer fast ​

Answer 1

Answer:

Consider △CAB and △CED

∠CAB=∠CED (given)

∠ACB=∠ECD (common angle)

So by AA similarity criteria , △CAB∼△CED

Hence proved.

Also corresponding sides are proportional in similar triangles.

So CECA=EDAB

CECD+DA=EDAB

108+7=x9

1015=x9

23=x9

x=6cm

Answer 2

Consider △CAB and △CED

∠CAB=∠CED (given)

∠ACB=∠ECD (common angle)

So by AA similarity criteria , △CAB∼△CED

Hence proved.

Also corresponding sides are proportional in similar triangles.

So CECA=EDAB

CECD+DA=EDAB

108+7=x9

1015=x9

X=6