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I'm trying to find the length of the segment from $H$ to $J$. The problem states the length of the segment from $B$ to $E$ is $494$, along with a couple ratios of the sides. $BC:CD = 2:3$ and $DE:EF:FG = 2:3:4$. The problem also makes it clear that $B, C$ and $D$ are collinear ; $D, E, F$ and $G$ are collinear ; $C, H$ and $G$ are collinear ; $B, H, J$ and $E$ are collinear and $C, J$ and $F$ are collinear, as seen in the diagram.

Since they give the length of $BE$, I believe I am supposed to find the ratio between $BE$ and $HJ$. However, they don't give any angles, nor relate the two given ratios together, so I'm doubting that there are any similar triangles or angle bisectors that could be used.

I remember reading that Ceva's theorem can be used to relate the ratios of sides, but upon further research, there should be three cevians and a point which they meet for the theorem to be used. Could some of the lines be extended to achieve this effect? Or is this even the right approach to the problem?

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Without loss of generality assume D to be origin and and the position vector of C $3\vec a$ and of B be $5\vec a$

E be $2\vec b $ , F be $5\vec b$ and G be $ 9\vec b$


Now assume CJ:JF = k:1-k and EJ :JB = l:1-l

J = ( k )$5\vec b$ + ( 1-k )$3\vec a$ = ( l )$5\vec a$ + ( 1-l ) $2\vec b$

Now coefficients of $\vec a$ and $\vec b$ are same

5k= 2(1-l) and 5l= 3(1-k)

K=$\frac{4}{19}$ and l is $\frac{9}{19}$

And hence CJ:JF is 4:15 , EJ:JB=9:10


EH:HB = n:1-n and CH:HG = m:1-m

H= ( n )$5 \vec a$ + ( 1-n ) $2\vec b$ = ( m )$9\vec b$ + ( 1-m ) $3\vec a$

5n= 3(1-m) and 2(1-n)=9m

n=$\frac{7}{13}$ and m=$\frac{4}{39}$

EH:HB = 7:6


EB = 494 implies HB = 228 , EJ = 234

HJ = 494-228-234 = 32

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In my opinion, the easiest approach here is to use barycentric coordinates. Let $X$ be a point on the line $DE$ such that $D-E-X$ and $DE:EX=2:\alpha$ (so, if $X=F$ then $\alpha=3$, and if $X=G$ then $\alpha=7$). Let $Y$ be the intersection of $BE$ and $XC$ (so, if $X=F$ then $Y=J$, and if $X=G$ then $Y=H$). Put $BY:YE=x:(1-x)$ and $XY:YC=y:(1-y)$; we aim to calculate $x$ (it is the ratio of $BY$ and $BE$).

Set barycentric coordinate system to be $D(1:0:0)$, $X(0:1:0)$ and $B(0:0:1)$. Since $BC:CD=2:3$, we have $C=\frac{3}{5}B+\frac{2}{5}D= (\frac{2}{5}:0:\frac{3}{5})$, and $DE:EX=2:\alpha$ implies $E=\frac{\alpha}{2+\alpha}D+\frac{2}{2+\alpha}X= (\frac{\alpha}{2+\alpha}:\frac{2}{2+\alpha}:0)$. Further, $BY:YE=x:(1-x)$ implies $Y=(1-x)B+xE=(\frac{\alpha}{2+\alpha}x:\frac{2}{2+\alpha}x:1-x)$, and $XY:YC=y:(1-y)$ implies $Y=(1-y)X+yC= (\frac{2}{5}y:1-y:\frac{3}{5}y)$.

So, we must have $\frac{2}{2+\alpha}x=1-y$ and $1-x=\frac{3}{5}y$, and by solving this system we obtain $x=\frac{4+2\alpha}{4+5\alpha}$.

Thus, for $\alpha=3$, $BJ:BE= \frac{10}{19}$, so $BJ=260$, and for $\alpha=7$, $BH:BE=\frac{18}{39}$, so $BH=228$. Now, $HJ=32$.

Alternatively, if you are not familiar with barycentric coordinates, you can calculate everything (more or less in the same way) in the coordinate system, say, $(D,\{\overrightarrow{DX},\overrightarrow{DB}\})$ (put $D(0,0)$, $X(1,0)$ and $D(0,1)$).

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Let $\mathcal{H}_{X,k}$ denote a homothety with center at $X$ and a extension factor $k$.

We have a following fact (Theorem):

If $\mathcal{H}_{M,k_1}$ and $\mathcal{H}_{N,k_2}$ are homotheties then their compostion $\mathcal{H}_{M,k_1}\circ \mathcal{H}_{N,k_2}$ is again some homothety $\mathcal{H}_{S,k}$ with $k=k_1k_2$ (if $k\ne 1$) and it center $S$ lies on a line $MN$.

Since we have: \begin{align} \mathcal{H}_{C, {-3\over 2}}: & \;B \longmapsto D\\ \mathcal{H}_{F,{3\over 5}}: &\; D \longmapsto E\\ \end{align} we see that $J$ is a center of homothety which takes $B$ to $E$ with ratio $-{9\over 10}$, so $BJ:JE = 10:9$.

Similary we get ${-7\over 6}$ for homothety from $H$ that takes $B$ to $E$, so $BH:HE = 6:7$.

Write $BX =y$, $HJ=x$ and $JE =z$ and you have a system of 3 equations to solve

\begin{align} x+y+z & = 494\\ 9x+9y &= 10z\\ 6x+6z &= 7y \end{align} and you are done.

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