Prove that the circle drawn with any one of the equal sides of an isosceles triangle as diameter bisects the unequal side.

Explanation

Let in `Delta ABC, AB = AC`,. The circle drawn with AB as a diameeter intersects BC at a point D.
To prove : D is the mid-point of BC, i.e, BD=CD
Proof: The circle intersect BC at a point `D :. angleADB` is a semi-circular angle.
`:. angle ADB=90^(@) rArr angle ADC= 90^(@)`
Again in `Delta ABC, AB = AC :. angle ABC= angleACB or, angle ABD= angle ACD`
So, in ` Delta ABD and Delta ACD, angle ADB= angle ADC [ :."each is " 90^(@)]`
`angle ABD= angle ACD and AD` is common to both.
`:. Delta ABD~= Delta ACD` [ by the A-A-S condition of congruency]
`:. BD=CD [ :. ` Similart sides of congruent triangle]
Hence D is the mid-point of BC, i.e., the circle bisects the unequal side.