Question

Prove that the circle drawn on any any one of the equal sides of an isosceles triangle and as diameter bisects the base ​

Answer 1

Answer:

let ABC is the isosceles triangle where AB = AC

The circle drawn on AC as diameter intersects BC at D

From Circle theorem we know that the angle inscribed on semicircle is 90 degree

Hence angle ADC is 90 degree

So we have angle ADB is 90 degree

This gives Δ ADC and Δ ADB are right angled triangle.

We have hypotenuse =AB=AC and AD common leg

Two right triangles are congruent if the hypotenuse and one corresponding leg are equal in both triangles.

Hence Δ ADC ΔADB are congruent

This gives BD=DC

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Answer 2

Answer:

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