Given: ∆ABD and ∆BCD and isosceles triangles on the same base BD.
To Prove: ∠ABC = ∠ADC
Proof: ∵ ∆ABD is isosceles
∴ AB = AD
∴ ∠ABD = ∠ADB ...(1)
| Angles opposite to equal sides of a triangle are equal
∵ ∆CBD is isosceles
∴ CB = CD
∴ ∠CBD = ∠CDB ...(2)
| Angles opposite to equal sides of a triangle are equal
Adding (1) and (2), we get,
∠ABD + ∠CBD = ∠ADB + ∠CDB
⇒ ∠ABC = ∠ADC