Prove that the altitude to the base of an isosceles triangle is also the angle bisector of the angle from which it is drawn
Please do statement and reasoning

Step-by-step explanation:

To prove- Altitude to the base of an isosceles triangle bisects the angle from which it is drawn.

In the given figure ΔABC (isosceles triangle) whose sides AB=AC

From ∠A an altitude is drawn to base (BC) which meets BC at X

∠AXB=90 and ∠AXC =90

∴We need to prove ΔXAB= ΔXAC for altitude to bisect the ∠ BAC

In Δ AXB and Δ AXC

AB=AC (same hypotenuse)

AX= common altitude (serve as the common leg of the hypotenuse)

ΔAXB ≅ ΔAXC

Hence, we can conclude that the corresponding angle∠ XAB = ∠XAC. QED

don't report but were is the digram

step-by-step explanation: