In the right-angled triangle ABC to the hypotenuse AB, the height CH is drawn. In the triangle ACH

In the right-angled triangle ABC to the hypotenuse AB, the height CH is drawn. In the triangle AСН, the median of the НM is drawn. Find the area of the triangle CHM if AC = p and the angle BAC is α.

Since CH is the height of the ABC triangle, the AСН triangle is also rectangular, and the median of the НM is drawn from the vertex of the right angle of the AНС to the hypotenuse of the AС. Then MH = AC / 2 = p / 2 cm.

Then the triangle CHM is isosceles in which CM = HM = p / 2 cm.

The AСM triangle is also isosceles, the angle MAH = AСM = α0. Then the angle AMН = (180 – 2 * α).

The СMН angle is adjacent to the AMН angle, then the СMН angle = (180 – 180 + 2 * α) = 2 * α.

Then the area of the triangle CHM will be equal to:

Scnm = CM * НM * SinCMH / 2 = (p / 2) * (p / 2) * Sin (2 * α) / 2 = (p ^ 2/6) * Sin (2 * α) cm2.