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Circle drawn on the side of an isosceles triangle

Posted January 31st, 2008 by alpha
in

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

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Answer

On January 31st, 2008 Isoscel says:

Let $ \triangle ABC $ be isosceles. AB=AC. let O be the middle of AB and $ \mathcal{C}(O,OA) $ the circle having AB as diameter.Let M be the intersection of $ \mathcal{C}(O,OA) $ with BC. Angle AMB is right as its measure is half of 180(measure of arc(AB)). Then $ \triangle AMB=\triangle AMC $ as are right triangles and AB=AC and AM is common.
Then BM=MC as the third pair of sides.

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