## [ [Mean proportionals in a right triangle]

[ Median to the hypotenuse ]

In a right triangle $A B C$, angle $A C B$ is a right angle. Let $E$ be the point of intersection of the angle bisector of angle $A B C$ with side $A C$. Point $D$ is the midpoint of side $A B$, and $O$ is the point of intersection of segments $B E$ and $C D$. A perpendicular is drawn from point $O$ to $B O$ until it intersects side $B C$ at point $F$. It is known that

$F C=b, O C=3 b / 2$. Find the area of triangle $A B C$.

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Draw the median $O L$ of triangle $O C F$.

## Solution

Draw the median $O L$ of the right triangle $O C F$. Then $O L=F L=L B, \angle O L B=2 \angle O B L=\angle B=$ $\angle B C D$. Therefore, $O F \| A B$, and triangle $C O F$ is isosceles (see figure). Consequently, $O F=O C=3 b / 2, B F$ $=3 b, B C=4 b, C L=5 b / 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_6409e68af5a1ba5e989dg-30.jpg?height=335&width=489&top_left_y=667&top_left_x=790)

By Thales' theorem, $C D={ }^{B C} / C L \cdot O C=12 b / 5, A B=2 C D=24 b / 5, A C^{2}=A B^{2}-B C^{2}=16 \cdot 11 b^{2} / 25, \quad S_{A B C}=1 / 2 B C \cdot A C=$ $\frac{8 b^{2} \sqrt{11}}{5}$

## Answer

$\frac{8 b^{2} \sqrt{11}}{5}$