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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 $\frac{\angle A_1MB_1}{\angle A_1CB_1}$.

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

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    @DavidQuinn Nope .. I think the answer would be independent of the side lengths – 2017-01-29
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    $\angle A_1CB_1 = 60^\circ$. Your angle chasing attempt was not too zealous. Try more, this one is solved by angle chasing. – 2017-01-30
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    @zhoraster Would you mind adding that solution ? – 2017-01-30

2 Answers 2

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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 $\angle A_1CB_1=\angle BC_1A_1$

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    Is $\angle A_1C_1B_1 = \angle A_1CB_1$? – 2017-01-29
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    I think it is $\pi-2\angle A_1CB_1$ – 2017-01-29
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    I added more information why that is true. – 2017-01-29
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    I think I said same thing... $\angle A_1C_1B_1 = \angle A_1CB_1$ – 2017-01-29
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    I hurried up and I haven't seen that $C$ is 60 degrees first time :) Sorry about that – 2017-01-29
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    Thanks :) Nice solution :) – 2017-01-29
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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

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