On the diagonals $A C$ and $B D$ of trapezoid $A B C D$ with bases $B C=a$ and $A D=b$, points $K$ and $L$ are located respectively, such that

$C K: K A=B L: L D=7: 4$. Find $K L$.

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Continue $K L$ until it intersects one of the lateral sides of the trapezoid.

## Solution

Draw a line through point $K$ parallel to the bases. Let $P$ and $L_{1}$ be the points of intersection of this line with side $C D$ and diagonal $B D$, respectively. From the theorem of proportional segments, it follows that $D P: P C = A K: K C = 4: 7, D L_{1}: L_{1} B = D P: P C = 4: 7$.

Therefore, point $L_{1}$ coincides with point $L$. Hence, $K L \| A D$.

From the similarity of triangles $C K P$ and $C D A$, we find that $K P = C K / A C \cdot A D = 7 / 11 A D = 7 / 11 b$, and from the similarity of triangles $D P L$ and $D C B - L P = 4 / 11 B C = 4 / 11 a$.

Therefore, $K L = |K P - L P| = |7 / 11 b - 4 / 11 a|$.

## Answer

$1 / 11 |7 b - 4 a|$.