[Theorem of the sum of squares of diagonals] [Two pairs of similar triangles]

In a triangle with sides 9 and 15, a parallelogram is inscribed such that one of its sides, equal to 6, lies on the third side of the triangle, and the diagonals of the parallelogram are parallel to the two given sides of the triangle. Find the other side of the parallelogram and the third side of the triangle.
The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides.

## Solution

Let the vertices $M$ and $N$ of the parallelogram $M N P Q$ be on the side $A C$ of triangle $A B C$, and the vertices $P$ and $Q$ on the sides $A B=9$ and $A C=15$.

Since $A P Q M$ and $N P Q C$ are parallelograms, then $A M=P Q=N C=6, A C=18$.

From the similarity of triangles $M Q C$ and $A B C$, it follows that $Q M=2 / 3 A B=6$, and from the similarity of triangles $A P N$ and $A B C$, $P N=2 / 3 B C=10$.

By the theorem on the sum of the squares of the diagonals of a parallelogram, $P N^{2}+M Q^{2}=2 P Q^{2}+2 P M^{2}$. Therefore, $P M^{2}=1 / 2\left(P N^{2}+M Q^{2}-2 P Q^{2}\right)=1 / 2(100+36-72)=32$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5ddd4eab4fe1a12edeb9g-11.jpg?height=369&width=612&top_left_y=1622&top_left_x=723)

Answer

$4 \sqrt{2}, 18$.