Question

If circles are drawn taking two sides of a triangle as diameters prove that the point of intersection of the circles lie on the third side.
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Answer 1

Answer:

A diameter of a circle divides the circle into 2 equal parts each  of these two equal parts is called a semicircle.

Angle in a semicircle is a right angle.

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Given,

Two circles are drawn with sides AB and AC of the triangle ΔABC as diameters. Both these circles intersect each other at D.

To Prove:

D lies on BC

Construction: Join AD

To prove:

Since,AC and AB are the diameters of the two circles.

∠ADB =90°......(i)

∠ADC = 90°......(ii)

 (Angle in the semi circle)

On adding eq i & ii

∠ADB + ∠ADC = 180°

∠BDC= 180°

Hence,BDC is straight line.

So , point of intersection D lies on the third side.

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Answer 2

$\huge \mathtt { \underline{SOLUTION:-}}$

Let ABC bi a triangle and two circles are drawn taking AB and AC as diameters.

Let D be the point of intersection of the two circles.

Let us assume, D does not lie on the third side.

Now,

=> <BDA = 90° and <CDA = 90° [Angle in a semicircle is a right angle]

Therefore,

=> <BDA+<CDA = 90°+90°

=> 180°

Therefore,

BDC is a straight line.

Therefore,

D lies on the third side BC

Hence proved ✔️