$ABC$
is isosceles with $AB=AC$
. Let the circle having diameter $AB$
and centre $O$
intersect $BC$
at some point $P$
. Find the ratio $\frac{BP}{BC}$
.

What can you say about the
$\angle APB$
? 
What about
$AP$
?

Angle
$\angle APB=\pi/2$
because it subscribes an arc length of$\pi$
(AB is a diameter), hence$AP$
is the height in the triangle. Being an isosceles triangle, this is also the median, and hence$\frac{BP}{BC}=\frac{1}{2}$
.Can you find an alternative proof considering the segment
$OP$
instead?