Question

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

Answer 1

Restatement

In the adjoining figure ABC is an isosceles triangle whose AB = AC. AD is the altitude of the ∆ABC. To show that AD bisects ∠BAC

Proof

Given ,

AB = AC

Since , the opposite pair of sides are equal the angles are also equal.

∴∠ACB = ∠ABC

$\sf Now \: from \: \triangle ADB \: and \: \triangle ADC \: we \: have :$

$\sf \bullet \: \: AB = AC$$\sf\bullet \: \: \angle ABD = \angle ACD \: \: \: \{ \angle ABC = \angle ACB \} \\\\ \sf\bullet \: \: AD = AD \: \: \: \{ Common \: side \}$

Thus , ∆ADB ≈ ∆ADC ( SAS )

$\sf \implies \angle BAD = \angle CAD$

Therefore , AD divides ∠BAC into two equal parts .

Hence , AD is bisects ∠BAC

Answer 2

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