If `A B C` is an isosceles triangle such that `A B=A C` and `A D` is an altitude from `A` on `B C` . Prove that (i) `/_B=/_C` (ii) `A D` bisects `B C` (iii) `A D` bisects `/_A`

Explanation

"Given: ABC is an isosceles triangle in which AB=AC.

To Prove: (i) `angleB=angleC`

(ii) AD bisects BC

(iii) AD bisects `angle A`

Proof : In `triangle ABD and ACD`

`AB=AC` (Given)

`AD=AD` (Common)

`triangle ADB=triangle ADC` (each `90^*`)

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