[ Trigonometric relationships in a right triangle ] Complex [ The ratio of the areas of triangles with a common base or common height] Class

In trapezoid $A B C D$, segments $A B$ and $C D$ are the bases. The diagonals of the trapezoid intersect at point $E$. Find the area of triangle $B C E$, if $A B=30, D C=24, A D=3$ and $\angle D A B=60^{\circ}$.
Find what fraction of the area of triangle $D C B$ the area of triangle $B C E$ constitutes.

## Solution

Let $D K$ be the height of the given trapezoid. From the right triangle $A K D$, we find that

$$
D K=A D \sin \angle D A B=\frac{3 \sqrt{3}}{2}
$$

The height of triangle $D C B$, drawn from vertex $B$, is also equal to $\frac{3 \sqrt{3}}{2}$. Therefore, its area is $\frac{3 \sqrt{3} \cdot D C}{4}$

$=18 \sqrt{3}$.

From the similarity of triangles $A E B$ and $C E D$, it follows that

$$
\frac{B E}{D E}=\frac{A B}{D C}=\frac{30}{24}=\frac{5}{4}
$$

Therefore, $\frac{B E}{B D}=\frac{5}{9}$. Consequently,

$$
S_{\triangle \mathrm{BCE}}=\frac{B E}{B D} \cdot S_{\triangle \mathrm{DCB}}=\frac{5}{9} \cdot 18 \sqrt{3}=10 \sqrt{3}
$$

## Answer

## Problem