if two circles are drawn taking two sides of a triangle as a diameter prove that the point of intersection of the circles lies on the third side
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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.
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Answer:
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.
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