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\ ABC is an isosceles triangle with A ...

`\ ABC` is an isosceles triangle with `A B=A C`. Side `B A` is produced to `D` such that `A B=A D`. Prove that `/_B C D` is a right angle.

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Given : `DeltaABC` in which `AB=AC`. Side BA is produced to D such that AB =AD.
To prove : `angleBCD=90^(@)` i.e., `angle2+angle3=90^(@)`
Proof : since `AB=AC` (given)
`:. angle2=angle1` ...(1)
(angle opposite to equal sides are equal)
Also, `AB=AC=AD` (given)
`angle3=angle4` (angles opposite to equal sides are equal) ...(2)
Now, in `DeltaBCD`,
`angle1+(angle2+angle3)+angle4=180^(@)` (angle sum property) ...(3)
`implies" "angle2+(angle2+angle3)+angle3=180^(@)` [from (1), (2) and (3)]
`implies" "2(angle2+angle3)=180^(@)`
`implies" "angle2+angle3=90^(@)`. Hence Proved.
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