To find the ratio of a portion of a diagonal of the parallelogram to the diagonal of the parallelogram given the following conditions.

Question

Consider a parallelogram $ABCD$. Suppose that $P$ is a point on the side $AD$ so that $AP : AD = x : y$ . Let $Q$ be the intersection point of $AC$ and $PB$. Show that $AQ : AC = x : x + y$ .

I tried to find similar triangles by constructing another diagonal. But couldn't proceed further.

Answer 1

$\triangle AQP\simeq \triangle CQB$. This can help you.