# In $\triangle CDF$, $CE$ $FB$ and $DG$ are placed in such a way that they intersects at point $H$. What is the value of $\frac{CH}{HE}$?

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.

• Is $A$ the center of the circle? – A gal named Desire Feb 7 at 18:22
• Yeah, but I didn't state that. – Anirban Niloy Feb 7 at 18:23
• 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. – A gal named Desire Feb 7 at 19:01
• If it was labeled on the diagram on the competition, you don't need to repeat the mistake in the diagram in your post. – A gal named Desire Feb 7 at 19:03

$${DB\over CB}\times{CG\over FG}\times{FE\over DE}=1 \implies {FE\over DE}={9\over 10}$$
$${CG\over FG}\times{FD\over ED}\times{EH\over CH}=1 \implies {EH\over CH}={15\over 19}$$
• 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, – cr001 Feb 7 at 17:14