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Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
Given:
Circles are described on the sides of a triangle as diameters.
To do:
We have to prove that the circles on any two sides intersect each other on the third side (or third side produced).
Solution:
Let in a $\triangle ABC$, circles are drawn on sides $AB$ and $AC$.
Draw $AD \perp BC$
$AD \perp BC$
This implies,
$\angle ADB = \angle ADC = 90^o$
From the figure,
The circles drawn on sides $AB$ and $AC$ as diameters will pass through $D$.
Hence the circles drawn on two sides of a triangle pass through $D$, which lies on the third side.
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