AD is an altitude of an isosceles DeltaA...

AD is an altitude of an isosceles `DeltaABC` in which AB = AC.
Show that (i) AD bisects BC,
(ii) AD bisects `angleA`.

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(i) In `ΔBAD` and `ΔCAD`,
`∠ADB = ∠ADC` (Each `90^@`as AD is an altitude)
`AB = AC` (Given)
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