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AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that (i) AD bisects BC (ii) AD bisects A.

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AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects A.

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It is given that AD is an altitude and AB = AC.

The diagram is as follows:

(i) In ΔABD and ΔACD,

ADB = ADC = 90°

AB = AC (It is given in the question)

AD = AD (Common arm)

∴ ΔABD ΔACD by RHS congruence condition.

Now, by the rule of CPCT,

BD = CD.

So, AD bisects BC

(ii) Again, by the rule of CPCT, BAD = CAD

Hence, AD bisects A.

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