Question

in fig, AC =AE, AB=AD and angle BAD = angleEAC. show that BC=DE(20 points)

Answer 1

In BAC and DAE
AC=AE (GIVEN)
AB=AD (GIVEN)
BAD+DAC = CAE+DAC (BAD=CAE)
BAC is congruent to DAE
BC=DE(CPCT)

Answer 2

Given : AC = AE, AB = AD, AND ∠BAD = ∠EAC

To prove : BC = DE

Proof : ∠BAD = ∠EAC ( given)

$\therefore \: adding \: \Angle \: DAC \: both \: side$

=> ∠BAD + ∠DAC = ∠EAC + ∠DAC

=> ∠BAC = ∠EAD

In triangle BAC and EAD

AB = AD ( Given)

AC = AE ( Given)

∠BAC = ∠EAD

$\therefore \: By \: SAS \: Congruence \: \triangle \: BAC \cong \: \triangle \: EAD$

Thus, by CPCT BC = DE

Proved