Question

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

Answer 1

Answer:

Solution:

First, draw a triangle ABC and then two circles having a diameter as AB and AC respectively.

We will have to now prove that D lies on BC and BDC is a straight line.

Proof:

As we know, angle in the semi-circle are equal

So, ∠ADB = ∠ADC = 90°

Hence, ∠ADB + ∠ADC = 180°

∴ ∠BDC is a straight line.

So, it can be said that D lies on the line BC.

Hope it will be helpful :)

Answer 2

Answer:

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Explanation:

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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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Hope it helps!!

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