On the diagonals $A C$ and $B D$ of trapezoid $A B C D$, points $M$ and $N$ are taken respectively such that $A M: M C=D N: N B=1: 4$. Find $M N$, if the bases $A D=a, B C=b(a>b)$.
Continue $M N$ until it intersects one of the lateral sides of the trapezoid.

## Solution

Draw a line through point $M$ parallel to the bases. Let $K$ and $N_{1}$ be the points of intersection of this line with side $C D$ and diagonal $B D$, respectively (see figure). From Thales' theorem, it follows that $D K: K C = A M: M C = 1: 4$, $D N_{1}: N_{1} B = D K: K C = 1: 4$. Therefore, point $N_{1}$ coincides with point $N$. Consequently, $M N \| A D$.

From the similarity of triangles $C K M$ and $C D A$, we find that $M K = 4 / 5 A D = 4 / 5 a$, and from the similarity of triangles $D K N$ and $D C B$, we find that $K N = 1 / 5 B C = 1 / 5 b$.

Therefore, $M N = M K - K N = 1 / 5 (4 a - b)$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-25.jpg?height=378&width=669&top_left_y=134&top_left_x=700)

Answer

$1 / 5 (4 a - b)$.

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