In trapezoid $A B C D(B C \| A D)$, the diagonals intersect at point $M, B C=b, A D=a$.

Find the ratio of the area of triangle $A B M$ to the area of trapezoid $A B C D$.
$S_{A B M}=B M /{ }_{B D} \cdot S_{A B D}, S_{A B D / S_{A B C D}}=a / a+b$.

## Solution

From the similarity of triangles $B M C$ and $D M A$, it follows that $B M: M D=B C: A D=b: a$, so ${ }^{B M} /{ }_{B D}=b / a+b, S_{A B M}=$ $B M /{ }_{B D} S_{A B D}=b_{/ a+b} S_{A B D}$

Let $h$ be the height of the trapezoid. Then $S_{A B C D}=1 / 2(a+b) h, S_{A B D}=1 / 2 a h, S_{A B D / S_{A B C D}}=a / a+b$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-42.jpg?height=88&width=1258&top_left_y=276&top_left_x=0)

## Answer

$a b /(a+b)^{2}$