In the adjoining figure, ABC is a triangle in which AD is the bisector of ∠A. If AD ⊥ BC, show that ∆ABC is isosceles.

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Solution

$\begin{array}{l} Given: \\ ∠ BAD = ∠ DAC ...(i) \\ To show that △ ABC is isoceles, we should show that ∠ B = ∠ C. \\ ∴ AD ⊥ BC, ∠ ADB = ∠ ADC = 90 °^{} \end{array} ∠ A D C = ∠ A D B ∠ B A D + ∠ A B D = ∠ D A C + ∠ A C D (e x t e r i o r a n g l e p r o p e r t y) ∠ D A C + ∠ A B D = ∠ D A C + ∠ A C D (from equation (i)) ∠ A B D = ∠ A C D This is because opposite angles of a triangle ∆ ABC are equal . Hence, ABS is an isosceles triangle .$

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