[ Product of chord segment lengths and product of secant segment lengths ] Law of Cosines

A circle passes through vertices $A$ and $C$ of triangle $A B C$, intersects side $A B$ at point $D$ and side $B C$ at point $E$. Find the angle $C D B$ if $A D=5, A C=2 \sqrt{7}, B E=4, B D: C E=3: 2$.

#
The product of the entire secant and its external part for a given point and a given circle is constant.

## Solution

Let $B D=3 x, C E=2 x$.

Since $B D \cdot A B=B E \cdot B C$, then $3 x \cdot(3 x+5)=4(2 x+4)$. From this, we find that $x=1$. Therefore,

$$
A B=A D+D B=8, B C=B E+E C=6
$$

Since $A C^{2}+B C^{2}=28+36=64=A B^{2}$, triangle $A B C$ is a right triangle, $\cos \angle B A C=\frac{A C}{A B}=\frac{\sqrt{7}}{4}$.

Using the cosine rule in triangle $A D C$, we find that

$$
\begin{aligned}
D C^{2} & =A D^{2}+A C^{2}-2 A D \cdot A C \cos \angle D A C= \\
& =25+28-2 \cdot 5 \cdot 2 \sqrt{7} \cdot \frac{7}{4}=18
\end{aligned}
$$

Using the cosine rule in triangle $B D C$, we find that

$$
\cos \angle B D C=\frac{B D^{2}+D C^{2}-B C^{2}}{2 B D \cdot D C}=\frac{9+18-36}{2 \cdot 3 \cdot 3 \sqrt{2}}=-\frac{\sqrt{2}}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b9b3326e3456e69a696eg-09.jpg?height=345&width=644&top_left_y=182&top_left_x=707)

## Answer

$\arccos \left(-\frac{\sqrt{2}}{4}\right)$