BdMO 2014 National Secondary Problem 6

Question

Let $\triangle ABC$ be an acute triangle with $\angle C = 60^{\circ}$. Perpendiculars $AA_1$ and $BB_1$ are drawn from point $A$ and $B$ to the sides $BC$ and $AC$ respectively. Let $M$ be the midpoint of $AB$. Find the value of $\frac{\angle A_1MB_1}{\angle A_1CB_1}$.

I tried angle chasing but it didn't help. How do I start?
Source: BdMO 2014 National secondary problem 6

Answer 1

Take the third altitude, $CC_1$. Then $A_1,B_1,C_1$ and $M$ are on the concyclic (on the nine point circle). S0 $\angle A_1MB_1=\angle A_1C_1B_1$ After this, it is easy:

$$\angle A_1MB_1=\angle A_1C_1B_1=180^\circ-2\angle A_1CB_1=60^{\circ}$$

EDIT: $B_1C_1BC$ is cyclic quadrilater ($\angle CB_1B=\angle CC_1B=90^\circ$), so $\angle A_1CB_1=\angle AC_1B_1$ and similarly $\angle A_1CB_1=\angle BC_1A_1$

Answer 2

There is another way to solve that:

You can also see that $A_1M$ is a median w.r.t hypotenuse of $\Delta AA_1B$ so

$$AA_1=BM\to \angle A_1MB=180º-2\angle B$$

We also have $B_1M$ is the median w.r.t hypotenuse of $\Delta AB_1B$ so

$$B_1M=AM \to \angle B_1MA=180º-2\angle A$$

Finally,

$$A_1MB_1=180º-(\angle A_1MB+\angle B_1MA)=2(\angle A+\angle B)-180º=180º-2\angle C=60º$$

and then

$$\frac{A_1MB_1}{A_1CB_1}=\frac{60º}{60º}=1$$