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Question
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution
Given that
∠BAD = ∠EAC
On adding ∠DAC on both sides, we get
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠EAD …(I)
Now, in △ABC and △AED,
AB = AD ...[Given]
AC = AE ...[Given]
∠BAC = ∠EAD ...[By (I)]
∴ △ABC ≌ △ADE ...[By AAS congruence rule]
⇒ BC = DE ...[Corresponding parts of congruent triangles]
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