Question

AB=AC , D is a point on AC and E on AB such that AD=ED=BC prove that angle AED = angle BCE​

Answer 1

Let the base angle of △ACD be ∠A.

∠D=180−2∠A

∠E=∠A

Consider △DEC

∠EDC=180−180−2∠A

⟹∠EDC=−2∠A

∠DEC=180−2∠A−2∠A=180−4∠A

∠CEB=180−180+4∠A−∠A

∴∠B=∠CEB=3∠A

∴

∠A

∠B

=

1

3

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