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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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Draw ABC as an ISOSCELes triangle. On the side AB (lateral side) mark the mid point O. Now as O as the center, draw a circle with radius = OB =OA. It may not intersect base of all isosceles triangles. But we choose base BC of our circle LONG enough so that it will intersect BC (base) at D.

Now, OB = OA = OD = radius. AB = 2 * radius = AC (isosceles triangle)

In triangle OBD, anle B = angle D as sides are equal. Since angle B = angle C, then angle B = angle C = angle D.

triangles OBD and ABC are similar. AB || OB, BD || BC. and angles are all equal.