In an acute-angled triangle ABC In an acute-angled triangle ABC, the angle A is 60 °, BB1 and CC1 are

In an acute-angled triangle ABC In an acute-angled triangle ABC, the angle A is 60 °, BB1 and CC1 are the heights of the triangle, and M is the midpoint of side BC. Prove that all sides of the triangle MB1C1 are equal.

Point M divides the hypotenuse of a right-angled triangle BCC1 in half, therefore it is the center of the circumscribed circle with a radius equal to BC / 2.

Then MC1 = BC / 2.

Similarly, point M bisects the hypotenuse of the right-angled triangle BB1C and B1M = BC / 2.

From two equalities it follows that MC1 = B1M and triangle C1B1M is isosceles, and <B1 = <C1.

<C1BM + <MCB1 = 180 ° – 60 ° = 120 ° (1);

<CB1M = <MCB1 (2);

<BC1M = <C1BM (3);

<BC1M + <MB1C = 120 ° (4);

We sum up all the angles of the triangles ВМС1 and В1МС. The total should be 360 °:

<C1BM + <BMC1 + <BC1M + <B1MC + <MCB1 + <CB1M = 360.