SOURCE: Bangladesh Math Olympiad 2014.
The inner circle of $\triangle CDF$ touches $CD$, $DF$ and $FC$ at $B$, $E$ and $G$ points respectively. $CE$, $FB$ and $DG$ meets at the point $H$. The side $CD$ is divided into $5:3$ ratio at the point $B$ and $CF$ is divided into $3:2$ ratio at the point $G$. What is the value of $CH:HE$?
My Attempt:
I was able to solve for the value of $CE$ with the help of trigonometry and all the length expressing by $x$, my calculation was that $CE$ $\approx$ $7.5216x$ (may be less or more) but I couldn't anyhow solve for the measurement of the length $HE$.
I will be very much gladful if anyone shows me how it can be solved with another method except trigonometry.
By Ceva Theorem,
${DB\over CB}\times{CG\over FG}\times{FE\over DE}=1 \implies {FE\over DE}={9\over 10}$
By Menelaus Theorem,
${CG\over FG}\times{FD\over ED}\times{EH\over CH}=1 \implies {EH\over CH}={15\over 19}$