Question

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

( i ) AB bisects BC

( ii ) AD bisects $\angle A$
​

Answer 1

Given:-

ABC is an isosceles △.

AB=AC

AD is altitude.

∠ADB=∠ADC=90°

To prove:-

(i) AD bisects BC, i.e., BD=CD

(ii) AD bisects ∠A, i.e., ∠BAD=∠CAD

Proof:-

In △ADB and △ADC,

∠ADB=∠ADC[Each 90°]

AB=AC[Given]

AD=AD[Common]

By R.H.S congruency,

△ADB≅△ADC

By C.P.C.T.

BD=CD

∠BAC=∠CAD

Hence proved.

Answer 2

$\huge \fbox \red{answer}$

Given:-

ABC is an isosceles △.

AB=AC

AD is altitude.

∠ADB=∠ADC=90°

To prove:-

(i) AD bisects BC, i.e., BD=CD

(ii) AD bisects ∠A, i.e., ∠BAD=∠CAD

Proof:-

In △ADB and △ADC,

∠ADB=∠ADC[Each 90°]

AB=AC[Given]

AD=AD[Common]

By R.H.S congruency,

△ADB≅△ADC

By C.P.C.T.

BD=CD

∠BAC=∠CAD

Hence proved.