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12/07/2021 11:55 am
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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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12/07/2021 11:56 am
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.