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

2
by jadeja

Log in to add a comment

by jadeja

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.

as OB = 1/2 AB , BD = 1/2 BC.

Hence the proof is done.

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.

as OB = 1/2 AB , BD = 1/2 BC.

Hence the proof is done.