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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.


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

Mathematics
RS Agarwal
Standard IX

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