$\left[\begin{array}{ll}\text { Ratios of areas } \\ \text { P } & \text { Product of lengths of chord segments and lengths of secant segments }]\end{array}\right]$ [ Ratio of areas of triangles with a common angle ]

In triangle $A B C$, points $B_{1}$ and $C_{1}$ are chosen on sides $A B$ and $A C$ respectively, such that $A B_{1}: A B=1: 3$ and $A C_{1}: A C=1: 2$. A circle is drawn through points $A, B_{1}$, and $C_{1}$. A line is drawn through point $B_{1}$, intersecting segment $A C_{1}$ at point $D$, and the circle at point $E$. Find the area of triangle $B_{1} C_{1} E$, if $A C_{1}=4, A D=1$, $D E=2$, and the area of triangle $A B C$ is 12.
Apply the theorem of intersecting chords.

## Solution

By the theorem of the product of segments of intersecting chords

$$
B_{1} D \cdot D E=A D \cdot D C_{1} .
$$

Therefore,

$$
B_{1} D=A D \cdot \frac{D C_{1}}{D E}=1 \cdot \frac{3}{2}=\frac{3}{2}
$$

Thus,

$$
\begin{gathered}
S_{\Delta \mathrm{B}_{1} \mathrm{C}_{1} \mathrm{E}}=\frac{B_{1} E}{B D} S_{\Delta \mathrm{B}_{1} \mathrm{DC}_{1}}=\frac{7}{3} S_{\Delta \mathrm{B}_{1} \mathrm{DC}_{1}}= \\
=\frac{7}{3} \cdot \frac{D C_{1}}{A C_{1}} S_{\Delta \mathrm{AB}_{1} \mathrm{C}_{1}}=\frac{7}{3} \cdot \frac{3}{4} S_{\Delta \mathrm{AB}_{1} \mathrm{C}_{1}}= \\
=\frac{7}{4} S_{\Delta \mathrm{AB}_{1} \mathrm{C}_{1}}=\frac{7}{4} \cdot \frac{A B_{1}}{A B} \cdot \frac{A C_{1}}{A C} S_{\Delta \mathrm{ABC}}=\frac{7}{4} \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot 12=\frac{7}{2} .
\end{gathered}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_0e468145bb05e34a17aag-15.jpg?height=292&width=620&top_left_y=1766&top_left_x=725)

## Answer

$\frac{7}{2}$