[ The ratio of the areas of triangles with a common base or a common height ] Complex [ The area of a figure is equal to the sum of the areas of the figures into which it is divided ] Class

On side $A B$ of triangle $A B C$, point $E$ is taken, and on side $B C$, point $D$ is taken, such that $A E=2$, and $C D=11$. Lines $A D$ and $C E$ intersect at point $O$. Find the area of quadrilateral $B D O E$, if $A B=B C=8$, and $A C=6$.

#
Find the ratio $\frac{A O}{O D}$.

## Solution

Let $S_{\Delta \mathrm{ABC}}=S$. Then

$$
S_{\triangle \mathrm{BEC}}=\frac{B E}{A B} \cdot S=\frac{3}{4} S
$$

Draw a line through vertex $A$ parallel to $B C$, and extend segment $C E$ to intersect this line at point $T$. From the similarity of triangles $T A E$ and $C B E$, we find that

$$
A T=B C \cdot \frac{A E}{B E}=\frac{8}{3}
$$

From the similarity of triangles $T A O$ and $C D O$,

$$
\frac{A O}{O D}=\frac{A T}{D C}=\frac{8}{3}
$$

Therefore,

$$
S_{\triangle \mathrm{CDO}}=\frac{D O}{A D} \cdot S_{\Delta \mathrm{ADC}}=\frac{3}{11} \cdot \frac{1}{8} \cdot S_{\Delta \mathrm{ABC}}=\frac{3}{88} S
$$

Thus,

$$
S_{\mathrm{BDOE}}=S_{\triangle \mathrm{BEC}}-S_{\Delta \mathrm{CDO}}=\frac{3}{4} S-\frac{3}{88} S=\frac{63}{88} S
$$

The height $B M$ of the isosceles triangle $A B C$ is found using the Pythagorean theorem in the right triangle $A B M:$

$$
B M=\sqrt{A B^{2}-A M^{2}}=\sqrt{64-9}=\sqrt{55}
$$

Then

$$
S=S_{\triangle \mathrm{ABC}}=\frac{1}{2} A C \cdot B M=3 \sqrt{55}
$$

Therefore,

$$
S_{\text {BDOE }}=\frac{63}{88} \cdot 3 \sqrt{55}=\frac{189 \sqrt{55}}{88}
$$

## Answer

$\frac{189 \sqrt{55}}{88}$.