Let $\overrightarrow {OA}$ and $\overrightarrow {OB}$ be two ray with common end point $O$.
Let $G$ be a point lying in the interior of the $\angle AOB$.
Construct a $\triangle OCD$ such that the sides $OC$ and $OD$ lie on rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$ respectively.
My approach:
$1$$\rightarrow$ Construct $OP$ which passes through $G$ such that $OG:GP=2:1$.
Now taking $P$ as center cut arcs on both rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$ with point of intersection being $M,N$ respectively.
$2$$\rightarrow$ Now draw a line segment $CD\parallel MN$ and passing through $P$ with both $C,D$ lying on rays $\overrightarrow {OA}$ and $\overrightarrow {OB}$.
Hence, $OCD$ is the required triangle.
Proof: Since $CD\parallel MN$ $\Rightarrow$$P$ is midpoint of $CD$ as $MN$ is chord of the circle with $P$ as center.
Question: This question came in one of our regional maths exams for $17$ marks but I got $3$ marks for this proof .
Am I wrong with the construction/proof anywhere? I want to confirm.
The 3 marks are for finding $P$. Now consider parallels through $P$ to the rays. They intersect the rays in two points, $M,N$. ($OMPN$ is a parallelogram.) Now build $C,D$ on the rays, so that $M,N$ are middle points for $OC$, $OD$.
($MN$ is thus mid line in $OCD$, $CD\| MN$, $P$ is in the middle of $CD$. Use the parallelogram $OMPN$ to show this.)