Question

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

Answer 1

in triangle ADB and ADC

AD = AD (COMMON SIDE)

ANGLE ADB = ANGLE ADC = 90°

(SINCE AD WAS ALTITUDE)

AB = AC (GIVEN)

BY RHS CRITERIA OF CONGRUENCE

TRIANGLE ADB ≈ TRIANGLE ADC

BY CORRESPONDING PARTS OF CONGRUENCE TRIANGLE, BD=BC

SO, WE CAN SAY AD BISECTS BC

HENCE PROVED