The center of a circle inscribed in an acute-angled isosceles triangle divides the height drawn

The center of a circle inscribed in an acute-angled isosceles triangle divides the height drawn to the base in a ratio of 5: 3. Find the radius of the circumscribed circle if the height drawn to the base is 32 cm

Let us denote the isosceles triangle given by the condition ABC, AB = BC, BH is the height, point O is the center of the inscribed circle.
According to the condition ВO / OH = 5/3, the whole height is 32 cm, one part is 32/8 = 4.
ВO = 20 cm;

OH = 12 cm – inscribed circle radius.
Draw the radius OP = OH to the BC side.
In the right-angled triangle BPO, we find the leg BP:
BP = √ (BO² – OP²) = √ (400 – 144) = √256 = 16 (cm).
The ВНС and BPO triangles are similar (acute angle B is common).
The similarity coefficient is:
ВН / BP = 32/16 = 2;
BC = 2 * ВO = 40 (cm);
HC = 2 * OR = 24 (cm).
The height of the ВН is the median in the triangle ABC, which means that
AC = 2 * HC = 48 (cm).
We find the area of ​​the triangle ABC:
S ABC = 1/2 * AC * BH = 768 (cm²).
The radius of the circumscribed circle is found by the formula:
R = a * b * c / 4S = 40 * 40 * 48/4 * 768 = 25 (cm).
Answer: 25 cm.