5. The bisectors $\mathrm{AD}$ and $\mathrm{BE}$ of triangle $\mathrm{ABC}$ intersect at point I. It turns out that the area of triangle ABI is equal to the area of quadrilateral CDIE. Find the greatest possible value of angle ACB.
Answer: $60^{\circ}$.

Solution. Let $\mathrm{S}(\mathrm{CDIE})=\mathrm{S}_{1}, \mathrm{~S}(\mathrm{ABI})=\mathrm{S}_{2}$, $S(B D I)=S_{3}, S(A I E)=S_{4}$ (see figure). Since the ratio of the areas of triangles with a common height is equal to the ratio of the bases, and the angle bisector divides the opposite side in the ratio of the adjacent sides, we have $\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\mathrm{CD} / \mathrm{BD}=$ $\mathrm{AC} / \mathrm{AB}$. Similarly, $\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right)$

$=A E / E C=A B / B C$. Since $S_{1}=S_{2}$, then

$\left(\mathrm{S}_{1}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{2}+\mathrm{S}_{3}\right)=\left(\mathrm{S}_{2}+\mathrm{S}_{4}\right) /\left(\mathrm{S}_{1}+\mathrm{S}_{3}\right)$, from which

$\mathrm{AB} / \mathrm{BC}=\mathrm{AC} / \mathrm{AB}$. Or $\mathrm{c}^{2}=\mathrm{ab}$. By the cosine theorem,

$\cos C=\left(a^{2}+b^{2}-c^{2}\right) /(2 a b)=\left(a^{2}+b^{2}\right) /(2 a b)-c^{2} /(2 a b)=\left(a^{2}+b^{2}\right) /(2 a b)-1 / 2 \geq 1-1 / 2=1 / 2$.

We obtained $\cos \mathrm{C} \geq 1 / 2$. Therefore, the measure of angle C does not exceed $60^{\circ}$.

Equality is achieved in an equilateral triangle.

Criteria.

The relationship between the sides $c^{2}=a b$ is obtained, but the estimate is not made or is incorrect: 2 points.

It is proven that the angle is no more than 60 degrees, but an example where the estimate is achieved is not provided: 6 points.