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

Explanation

Given: `/_\ABC` is an isosceles triangle with `AB=AC`. A circle is drawn taking `AB` as diameter which intersects the side `BC` at `D`
To prove: `BD=DC`
Construction: Join `AD`
Proof: `/_ADB=90^@`[Angle in a semi-circle is right angle]
`/_ADB+/_ADC=180^@`
`=>/_ADC=90^@`
In `/_\ABD` and `/_\ACD`
`AB=AC` [Given]
`/_ADB=/_ADC`
`AD=AD` [Common]
`:./_\ABD~=/_\ACD` [`RHS` congruence criterion]
`therefore BD=DC` [By `CPCT`]
Hence Proved.