Question

In triangle ABC, the bisector AD of angle A is perpendicular to side BC Show that AB = AC and triangle ABC is isosceles.​

Answer 1

Step-by-step explanation:

in ∆ABD and ∆ACD..

DAB=DAC (AD is a bisector of angle A)

ADB =ADC=90 (AD is a perpendicular)

AD=AD (common side in both∆)

By ASA congruence rule ∆ABD congruent to ∆ACD

AB=AC ( congruent part of congruent triangles)

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Answer 2

$in \: ∆abd \: and \: ∆acd.. \\ < bad \: = \: < cad \: \: \: \: \: \: \: (given)$

$ad \: = ad \: \: \: \: \: (common)$

$< adb = \: < adc \: = {90}^{o} \: \: (given)$

So , ∆ABD ≈ ∆ACD. (ASA rule)

So,. AB = AC. (CPCT)

or, ∆ABC is an Isoceles Triangle

Hence Proved