△$ABC$ with $∠C = 60°$, heights $AA_1$ and $BB_1$ and midpoint $M$ on $BC$. show that $∆A_1 B_1 M$ is equilateral

Question

Assume that △$ABC$ is an acute triangle where $∠ACB = 60°$. Let $A_1$ and $B_1$ be two heights of the triangle and let $M$ be in the centre of the side $AB$. How to show that $∆A_1 B_1 M$ is an equilateral triangle (all sides are equal)?

Answer 1

Well, first of we know that the median from the right angle to the hypotenuse in a right triangle is half of the hypotenuse. Therefore $MB_1=\frac{AB}{2}=MA_1$. Also, we know that if $CC_1$ is the last height of the triangle then $\angle A_1MB_1=\angle A_1C_1B_1$ since points $A_1,B_1,C_1,M$ lie on a circle . However, since points $A,B_1, H, C_1$ and $A_1,B, H, C_1$ also lie on circles we can see that $\angle A_1MB_1=\angle A_1C_1B_1 = \angle A_1C_1H + \angle HC_1B_1 = \angle CAA_1 + \angle CBB_1 = 30^\circ+30^\circ=60^\circ$. Therefore triangle $MA_1B_1$ is equilateral.

Answer 2

Given that $\angle AB’B=\angle AA’B =90$ and the midpoint M, ABA'B' are cyclic with M being the center of the circle. Then, B'M = A'M and

$$\angle A'MB' =2\angle CAA' = 2(90-\angle C) = 60$$

Thus, the triangle MA'B' is equilateral.