3. The lateral sides $K L$ and $M N$ of trapezoid $K L M N$ are equal to 15 and 12, respectively, and the base $L M=3$. The bisector of angle $N K L$ passes through the midpoint of side $M N$. Find the area of the trapezoid. Answer: 80.
# Solution.

![](https://cdn.mathpix.com/cropped/2024_05_06_8490c75cc8eda7f101fag-2.jpg?height=236&width=638&top_left_y=1550&top_left_x=236)

Let $Q$ be the midpoint of segment $M N$. Extend the bisector of angle $N K L$ to intersect line $L M$ at point $P$. $\angle N K Q = \angle Q K L$.
$\angle N K Q = \angle K P L$ (alternate interior angles). Then $\angle Q K L = \angle K P L$ and $\triangle K P L$ is isosceles. $K L = L P$. Therefore, $M P = K L - M L = 12$.

$\triangle K Q N = \triangle M P Q (M Q = Q N, \angle K Q N = \angle M Q P, \angle K N Q = \angle Q M P)$.

Then $K N = M P = 12$.

Draw $M R \parallel K L (R \in K N)$. $M R = K L = 15$, $M N = 12$, $R N = K N - M L = 12 - 3 = 9$, $12^2 + 9^2 = 15^2$. Therefore, $\angle M N R = 90^\circ$ and $M N$ is the height of the trapezoid.

$\mathrm{S}_{K L M N} = \frac{1}{2}(K N + M L) \cdot M N = \frac{1}{2}(3 + 12) \cdot 12 = 80$.

Recommendations for checking.

Base $K N$ found: 4 points.

Height of the trapezoid found: 3 points.

Arithmetic error: -1 point.