In rhombus $ABCD$, heights $BP$ and $BQ$ intersect diagonal $AC$ at points $M$ and $N$ (point $M$ lies between $A$ and $N$), $AM=$ $p, MN=q$. Find $PQ$.

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Triangles AMP and CMB are similar, as are triangles $B M N$ and $B P Q$.

## Solution

The rhombus is symmetric with respect to the diagonal $B D$. Therefore, $P Q \| A C$ and $C N=A M$. Triangles AMP and CMB are similar by two angles, so ${ }^{B M} / M P={ }^{C M} / A M=p+q / p$.

Triangles $B M N$ and $B P Q$ are also similar, so $P Q / M N={ }^{B P} /{ }_{B M}=A C / M C=2 p+q / p+q$. Therefore, $P Q=2 p+q / p+q \cdot q$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4c124ad32a76c9faf91dg-25.jpg?height=320&width=512&top_left_y=843&top_left_x=773)

## Answer

$q(2 p+q) / p+q$