triangle problems

Question

So, I was sent a problem by someone. I solved it intuitively but couldn't come up with a logical answer as I got stuck at 1 step.

This is how I approached the problem - I drew a side equal in length to create an equilateral triangle coincidentally making an isosceles triangle with the adjoining line segments $MB$ and $AC$. Now I extended the line going from $C$ till it touched the side $AB$. The problem is that when I assume that line to be parallel to the segment $MB$ everything fits in perfectly but I've spent hours trying to prove lines $OP$ and $MB$ to be parallel. Can someone please tell me how I can approach at it to prove that $MB$ and $OP$ parallel.

Answer 1

Let $AB=AC=1$. By applying the sine rule to $\Delta AOB,\Delta AOC$ we get $AO=\frac{1}{\sin 150^\circ} \sin 10^\circ, \frac{AO}{\sin(120^\circ-x)}=\frac{1}{\sin x}$, which yeilds $x=\frac{5\pi}{9}=100^\circ$, hence the desired parallelity.

Answer 2

Important Comment: I messed the labelings, my point A is your C and my point C is your B. In the following I use the labelings as in my picture.

One way to do it is by trig. Let $N$ be the intersection of $OM$ and $AC$. One way to prove that $AO$ and $MC$ are parallel is by showing that $ON/NM=AN/NC$. We can compute these quantities by using law of sines. $$\frac{ON}{NM}=\frac{ON}{NC}\cdot \frac{NC}{NM}=\frac{\sin 40}{\sin 30}\cdot\frac{\sin 80}{\sin 30}$$ and $$\frac{AN}{NC}=\frac{AN}{NM}\cdot \frac{NM}{NC}=\frac{\sin 60}{\sin 10}\cdot\frac{\sin 30}{\sin 80}$$ Let's start with $\frac{\sin 60}{\sin 10}\cdot\frac{\sin 30}{\sin 80}$ and treat it to convert it in the other expression. \begin{align}\frac{AN}{NC}&=\frac{\sin 60}{\sin 10}\cdot\frac{\sin 30}{\sin 80}\\ &=\frac{\sin 60}{\sin 10}\cdot\frac{1/2}{\cos 10}\\ &=\frac{\sin60}{2\sin 10\cos 10}\\ &=\frac{\sin60}{\sin 20} \end{align} Now, I am going to use the following (useful) formula: $$\sin(3\alpha)=4\sin(\alpha)\sin(60+\alpha)\sin(60-\alpha)$$ In particular, for $\alpha=20$, we have that $\sin(60)=4\sin(20)\sin(80)\sin(40).$ Hence, \begin{align} \frac{AN}{NC}&=\frac{\sin 60}{\sin 20}\\ &=\frac{4\sin(20)\sin(80)\sin(40)}{\sin 20}\\ &=\frac{\sin 40\sin 80}{(1/2)^2}\\ &=\frac{\sin 40}{\sin 30}\cdot\frac{\sin 80}{\sin 30}=\frac{ON}{NM} \end{align} And we are done.