[Theorem of the tangent and secant; the product of the entire secant and its external part [

Area of a triangle (using height and base)

The center of the circle touching side $B C$ of triangle $A B C$ at point $B$ and passing through point $A$ lies on segment $A C$. Find the area of triangle $A B C$, given that $B C=6$ and $A C=9$.
Apply the tangent-secant theorem.

## Solution

Let $R-$ be the radius of the circle. Then

\[
(AC - 2R) AC = BC^2
\]

From this, we find that $R = \frac{5}{2}$.

Let $H$ be the foot of the perpendicular from point $A$, and $O$ be the center of the circle. Then \(\frac{AH}{OB} = \frac{AC}{CO}\). Therefore,

\[
AH = \frac{AC \cdot OB}{CO} = \frac{AC \cdot R}{AC - R} = \frac{45}{13}
\]

Thus,

\[
S_{\triangle \mathrm{ABC}} = \frac{1}{2} BC \cdot AH = \frac{135}{13}
\]

![](https://cdn.mathpix.com/cropped/2024_05_06_16ab69fab1743849dba9g-10.jpg?height=349&width=600&top_left_y=20&top_left_x=732)

## Answer

\(\frac{135}{13}\).