Question

prove that the circle drawn on any of the equal sides of a isosceles triangle as diameter bisects the third side

Answer 1

Let the ΔABC be our basis of drawing
Given-AB=AC 
           AB and AC are diameters of circles with center O and O' respectively 
          The circles intersect at A and D
To prove-BD=DC
Proof-
By Pythogoras's theorem,
Hypotenus² = SideA²+SideB²
∴In ΔADB,
AB²=AD²+BD²  (i)

Similarly, in ΔADC,
AC²=AD²+DC² (ii)

∵AC=AB
∴AC²=AB²

∴On equating (i) and (ii)
AD²+BD²=AD²+DC²
⇒ BD²=DC²
⇒ BD=DC

∴D bisects BC. 
Hence proved

P.S- the figure is attached.