I have this question on my "History of Mathematics" problem set:
Draw any angle $\angle AOB$; Pick a point $C$ on $OB$; Now trace $CD$ which is perpendicular to $OA$; Draw a parallel line $s$ to $OA$ which passes through $C$; Now trace the segment $OQ$, with $Q$ in $s$, which intersects $CD$ at $P$ in a way that $PQ = 2OC$.
Prove $\angle AOQ$ is a third of $\angle AOB$.
I have mande the draw, but I have no idea how to get through the proof
Any help is quite welcome! Thanks in advance.
Call $b = m(AOB)$ and $a = m(POC)$. Then
$\frac{OC}{\sin OPC} = \frac{PC}{\sin{POC}} = \frac{PQ\sin{PQC}}{\sin{POC}}$
Write the angles in terms of $a$ and $b$ and you'll see why $a = \frac{2}{3}b$, hence $AOQ$ is a third of $AOB$.