Prove the relation $\mathsf{H}(AB,CH)$

Question

$AC$ is a line segment with a point $B$ on it such that $|AB|=12,|BC|=3$. $CD$ is any line segment through $C$, where $D$ lies outside $AC$. Let $E$ be a point on $CD$. Let $BD$ and $AE$ intersect at $F$, and $AD$ and $BE$ intersect at $G$. If $GF$ intersects $AC$ at $H$, we need to prove $$|AH|=10.$$

Answer 1

Using Menelaus' theorem on $\triangle ABG$ with $\overline{DEC}$ as transversal, we have $$\begin{aligned}\dfrac{|AC|}{|CB|}\cdot \dfrac{|BE|}{|EG|}\cdot\dfrac{|GD|}{|DA|}&=1 \ (\text{taking magnitudes only}) \\ \implies \dfrac{|BE|}{|EG|}\cdot\dfrac{|GD|}{|DA|} &= \dfrac{|CB|}{|AC|}\\&=\dfrac3{15}=\dfrac{1}{5} \qquad (1) \end{aligned}$$ and using Ceva's theorem on $\triangle ABG$ with cevians $AE,BD,GH$, we have $$\begin{aligned} \dfrac{|AH|}{|HB|}\cdot \dfrac{|BE|}{|EG|}\cdot\dfrac{|GD|}{|DA|}&=1\\ \implies \dfrac{|AH|}{|HB|}\cdot\dfrac15&=1 \ \ (\text{from }(1))\\ \implies |AH|&=5|HB|\\ \implies |AH|+|HB|&=6|HB|\\ \implies |AB|&=6|HB|\\ \implies |HB|&=2 \ \ (\because |AB|=12) \\ \implies |AH|=|AB|-|HB|&=10\end{aligned}$$

Answer 2

From the construction we can see that $(ABCH)$ is harmonic. It means that $\frac{AC}{CB}:\frac{AH}{HB}=-1$. From this we get that $AH=5HC$, thus $H$ divides $AC$ in ratio $5:1$ which proves that $AH=10$.