If from the intersection point of the angle bisectors of an isosceles triangle, we draw perpendiculars to the sides, the three perpendiculars divide the triangle into two smaller and one larger kite. For which isosceles triangle will the sum of the areas of the two smaller kites be equal to the area of the larger kite?

![](https://cdn.mathpix.com/cropped/2024_05_02_3f3ea0d7f48f4e859f80g-1.jpg?height=271&width=225&top_left_y=306&top_left_x=945)
Solution. The intersection point of the angle bisectors is the center of the inscribed circle, and the perpendiculars erected to the sides from this point are the radii of the inscribed circle. Based on this, using the notations in the diagram, the area of the large deltoid is $a_{1} \cdot r$, and the area of the small deltoids is $a_{2} \cdot r$. Since the two smaller deltoids are congruent, and according to the condition, the area of the small deltoids is half the area of the large deltoid, thus $2 a_{2}=a_{1}$. The base of the triangle is $2 a_{2}$, and its legs are $a_{1}+a_{2}=3 a_{2}$ in length. Therefore, the sought triangles are those isosceles triangles that are similar and have a base length that is $\frac{2}{3}$ of the length of their legs.