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Prove that the circle drawn on any one o...

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

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Given : `A` `DeltaABC`in which `AB=AC` and a circle is drawn by taking AB as diameter which intersects the side BC of triangle at D.
To prove : `BD=DC`
Construction: Join AD.
Proof: Since angle in a semi-circle is a right angle, therefore
`angle ADB=90^@`
Now, in `DeltaABD` and `Delta ACD`, we have
`AB=AC` (given)
`angle ADB=AD` (each equal to `90^@`)
and, `AD=AD` (Common)
`rArrDeltaABDcongDelta ACD` (by `SAS` congruency)
`rArrBD=DC`
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