Question

AD is an altitude of an isosceles triangle ABC in which AB = AC, show that. i. AD bisects BC​

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.