The base of an isosceles triangle is $4 \sqrt{2}$, and the median drawn to the lateral side is 5.

Find the lateral sides.
Complete the given triangle to a parallelogram.

## Solution

Let $x$ be the side $A B$ of the isosceles triangle $A B C (B C=4 \sqrt{2})$. On the extension of the median $B M$ beyond point $M$, we lay off the segment $D M$ equal to $B M$. Then $B A D C$ is a parallelogram. Therefore,

$$
A C^{2}+B D^{2}=2\left(A B^{2}+B C^{2}\right), \text { or } x^{2}+10^{2}=2(4 \sqrt{2})^{2}+2 x^{2}
$$

From this, we find that $x^{2}=36$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-32.jpg?height=386&width=778&top_left_y=430&top_left_x=640)

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