# A triangle ABC is inscribed in the circle, the side of which AC coincides with the diameter.

**A triangle ABC is inscribed in the circle, the side of which AC coincides with the diameter. From point B to AC, a perpendicular BK is drawn, with AK = 4, and KC = 16. Find: BK, AB, BC, AC.**

The inscribed angle B of the triangle ABC rests on the arc AC = 180 °, which means that the angle B itself is equal to AC / 2 = 90 °, that is, the angle B is right, and the triangle ABC is right-angled.

Let’s omit the height BK. By the property of the height of a rectangular (!) Triangle, lowered to the hypotenuse, we have: BK² = AK * CK, hence BK = √ (4 * 16), BK = 8.

By the Pythagorean theorem, we find AB: AB² = BK² + AK², AB = √ (64 + 16) = 4√5.

In the same way we find BC: BC² = BK² + KC², BC = √ (256 + 64) = 8√5.

AC is found by the problem statement: AC = AK + CK = 4 + 16 = 20.