4.2. Let $E$ be the point of intersection of the diagonals of a convex quadrilateral $A B C D$. It is known that the perimeters of triangles $A B E, B C E, C D E, D A E$ are equal, and the radii of the inscribed circles of triangles $A B E, B C E, C D E$ are $3,4,6$ respectively. Find the radius of the inscribed circle of triangle $D A E$.
4.2. Answer: $\frac{9}{2}$.

Let the angle between the diagonals be $\alpha$. Then $S(A B E)=\frac{1}{2} A E B E \sin \alpha, S(B C E)=\frac{1}{2} B E C E \sin \alpha, S(C D E)=\frac{1}{2} C E D E \sin \alpha$, $S(D A E)=\frac{1}{2} D E A E \sin \alpha$. Notice that from the above formulas, it follows that $S(A B E) S(C D E)=S(B C E) S(D A E)$. Additionally, we know that $S=\frac{p r}{2}$, where $r$ is the radius of the inscribed circle, and $p$ is the perimeter of the triangle. Using this formula, we get

$$
r(A B E) p(A B E) r(C D E) p(C D E)=r(B C E) p(B C E) r(D A E) p(D A E)
$$

According to the problem, the perimeters of all triangles are equal, so $r(A B E) r(C D E)=r(B C E) r(D A E)$ and $r(D A E)=\frac{r(A B E) r(C D E)}{r(B C E)}$ $\frac{9}{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_04_b9e7a1c03f5c15132d5ag-18.jpg?height=974&width=988&top_left_y=67&top_left_x=543)

Fig. 12: to the solution of problem 3.2