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Question

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

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Solution

Given,

Two circles are drawn on the sides AB and AC of the triangle

ABC as diameters. The circles intersected at D.

Construction: AD is joined.

To prove: D lies on BC. We have to prove that BDC is a straight line.

Proof:

ADB=ADC=90° ...Angle in the semi circle

Now,

ADB+ADC=180°

BDC is straight line.

Thus, D lies on the BC.

490363_464053_ans_2fa9113e1b974e009c69c34dc2352ab8.PNG

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