Let $CH$ be the height of the triangle, $K, L$ be the points of intersection of the line passing through $C$ parallel to $AB$ with $AF$ and $BF$ respectively (see figure). Since the trapezoid $AKLB$ is circumscribed around a circle with diameter $CH$, $KD$ and $LD$ are the angle bisectors of $\angle AKL$ and $\angle BLK$ respectively. Therefore, $\angle CKD = 90^{\circ} - \angle HAD$, which means triangles $KCD$ and $DHA$ are similar, and $KC = CD^2 / AH = CH^2 / 4AH = BH / 4$. Similarly, $CL = BH / 4$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-36.jpg?height=515&width=803&top_left_y=646&top_left_x=639)

Therefore, the ratio of the heights of similar triangles $FKL$ and $FAB$ is $1: 4$, and the ratio of the heights of triangles $AFB$ and $ABC$ is $4: 3$.

## Answer

## $4: 3$.
[ Proportional means in a right triangle ] [ Isosceles, inscribed, and circumscribed trapezoids

In a right triangle $A B C$, point $D$ is the midpoint of the height dropped to the hypotenuse $A B$. Lines symmetric to $A B$ with respect to $A D$ and $B D$ intersect at point $F$. Find the ratio of the areas of triangles $A B F$ and $A B C$.

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