Question

prove that a circle drawn on one of the equal side of isosceles triangle as diameter bisects the base of the triangle. ​

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