Question

In figure AC =AE,AB=AD and traingle BAD= traingle EAC. show that BC=DE

Answer 1

given
AC=AE
AB=AD
angle BAD=angle CAE

to prove
BC=DE

Proof
angle BAD=angle CAE
Angle BAD+ angleDAC=angle CAE+
angleDAC
=angle BAC=angle DAE

IN triangleABC and triangle DAC
AB=AD
AC=AE
angle BAC=angle DAE
triangle ABC is congruent to triangle DAE
(by sas similarly criteria)
therefore BC=DE(bycpct)

Answer 2

 

Given,

AC = AE, AB = AD and ∠BAD = ∠EAC

 

To prove:

BC = DE

Proof: We have

∠BAD =

∠EAC

(Adding ∠DAC to both sides) ∠BAD +∠DAC =∠EAC +∠DAC

⇒ ∠BAC = ∠EAD

In ΔABC and ΔADE,

AC = AE (Given)

∠BAC =∠EAD (proved above)

AB = AD       (Given)

Hence, ΔABC ≅ ΔADE   (by SAS congruence rule)

Then,

BC = DE        ( by CPCT.)