Question

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

Answer 1

Step-by-step explanation:

Let's consider the two triangles △ABD and △ACD.

In △ABD and △ACD,

∠ADB=∠ADC (each 90° as AD is altitude)

AB=AC (given)

AD=AD (Common side)

Therefore, △ABD≅△ACD (By RHS criterion)

=>BD=CD

Hence D is the midpoint of BC.

So, AD bisects BC. (Proved)