In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

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Solution

It is given that ∠BAD = ∠EAC

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

In ΔBAC and ΔDAE,

AB = AD (Given)

∠BAC = ∠DAE (Proved above)

AC = AE (Given)

∴ ΔBAC ≅ ΔDAE (By SAS congruence rule)

∴ BC = DE (By CPCT)

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In the given figure, AC = AE, AB = AD and $∠ B A D = ∠ E A C .$ Show that BC = DE.

A C = A E, A B = A D

∠ B A D = ∠ E A C

B C = D E

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