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

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Solution

$\begin{matrix} A B = A C (g i v e n) ∴ ∠ A B C = ∠ A C B = θ A D a l t i t u d e T h e n, ∠ A D B = ∠ A D C = 90^{0} H e n c e ∠ B A D = ∠ D A C A D b i s e c t ∠ A \end{matrix}$

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

(i) AD bisects BC (ii) AD bisects ∠A.

Question 2 (ii)
AD is an altitude of an isosceles triangles ABC in which AB = AC.
Show that AD bisects $∠ A$ .

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