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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.

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  • $\begingroup$ Is $A$ the center of the circle? $\endgroup$ – A gal named Desire Feb 7 at 18:22
  • $\begingroup$ Yeah, but I didn't state that. $\endgroup$ – Anirban Niloy Feb 7 at 18:23
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    $\begingroup$ You should not label the center of the circle, especially when another point that is labeled and should be labeled is so close to it. $\endgroup$ – A gal named Desire Feb 7 at 19:01
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    $\begingroup$ If it was labeled on the diagram on the competition, you don't need to repeat the mistake in the diagram in your post. $\endgroup$ – A gal named Desire Feb 7 at 19:03
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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,

${CG\over FG}\times{FD\over ED}\times{EH\over CH}=1 \implies {EH\over CH}={15\over 19}$

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    $\begingroup$ Ceva Theorem states if three lines from vertices of a triangle meets at a center point, Then the product of the ratios of the divided sides, clockwisely(or counterclosewisely) multiplies to one. Menelaus Theorem states if a line cuts a triangle $ABC$ and intersects $AB,BC,CA$ at $X,Y,Z$ then ${AX\over BX}\times{BY\over CY}\times{CZ\over AZ}=1$. If you are doing contest math, both Theorem should bee practiced and familiarized as they appear A LOT in national level contest problems, $\endgroup$ – cr001 Feb 7 at 17:14
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    $\begingroup$ Thank you for your kind information and instruction. I am a new beginner. So, I ain't familiar with these theorem but I will try to be acquainted with those theorem. $\endgroup$ – Anirban Niloy Feb 7 at 17:19

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