In an isosceles triangle $A B C$, points $M$ and $N$ are located on the lateral sides $A B$ and $B C$ respectively.

Find the area of triangle $A B C$, given that $A M=5, A N=2 \sqrt{37}, C M=11, C N=10$.
Drop perpendiculars $M P$ and $N Q$ to the base $A C$; using the Pythagorean theorem, form an equation in terms of $x=A P$.

## Solution

Let $P$ and $Q$ be the projections of points $M$ and $N$ onto $A C$. Denote $A P=x$. Then, from the similarity of triangles $N Q C$ and $M P A$, we get that $Q C=2 x$.

$N C^{2}-C Q^{2}=A N^{2}-A Q^{2}$. Therefore, $A Q^{2}=148+4 x^{2}-100=48+4 x^{2}$.

Similarly, $C P^{2}=96+x^{2}$. Since $C P-A Q=(A C-x)-(A C-2 x)=x$, we have $\sqrt{96+x^{2}}-\sqrt{48+4 x^{2}}=x$. Solving this equation, we get $x=2$. Thus, $C P=10, A C=12, B C=15$. Using Heron's formula, $S_{A B C}=\sqrt{21 \cdot 6 \cdot 6 \cdot 9}=18 \sqrt{21}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3b3613da627e706c6878g-34.jpg?height=446&width=535&top_left_y=0&top_left_x=767)

## Answer

$18 \sqrt{21}$.