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  • 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.

Q 10. 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.

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Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

Proof: $A B$ is the diameter of the circle and $\angle A D B$ is formed in a semi-circle.
$\angle A D B=90^{\circ}$-------1 (angle in a semi-circle)
Similarly,
$A C$ is the diameter of the circle and $\angle A D C$ is formed in a semi-circle.
$\angle A D C=90^{\circ}$------2(angle in a semi-circle)
From 1 and 2, we have
$\angle \mathrm{ADB}+\angle \mathrm{ADC}=90^{\circ}+90^{\circ}=180^{\circ}$

$\angle A D B$ and $\angle A D C$ are forming a linear pair.

Hence, point D lies on this side.

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