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

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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?

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