The inscribed semicircle has its center at the midpoint of th base. It is tangent to the side of the isosceles triangle, so a radius makes a 90° angle there.
The long side of the isosceles triangle can be found from the Pythagorean theorem to be ...
BC² = BD² +CD²
BC² = 8² +15² = 289
BC = √289 = 17
The radius mentioned (DE) creates right triangles that are similar to ∆BCD. In particular, we have ...