# Q: 10    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.

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

Proof: AB is the diameter of the circle and $\angle$ADB is formed in a semi-circle.

$\angle$ADB = $90 \degree$........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ADC is formed in a semi circle.

$\angle$ADC = $90 \degree$........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ADB+$\angle$ADC=$90 \degree$+$90 \degree$=$180 \degree$

$\angle$ADB and $\angle$ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

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