Prove that the altitude to the base of an isosceles triangle is also the angle bisector of the angle from which it is drawn
1 answer:
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
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