Quadrilateral extensions

We place the configuration on a coordinate system, so that point $E$ is at the origin and points $A$ and $B$ lie on the positive $x$ -axis. Let $A\left( a,0 \right)$ , $B\left( b,0 \right)$ and $C\left( c,d \right)$ . Then we have $\overrightarrow{EA}=\left( a,0 \right) \ \ \ \ \ \overrightarrow{EB}=\left( b,0 \right) \ \ \ \ \ \overrightarrow{EC}=\left( c,d \right)$ Since $\overrightarrow{EC}$ and $\overrightarrow{ED}$ have the same direction, there exists a positive number $p$ , such that $\overrightarrow{ED}=p\overrightarrow{EC}$ , i.e. $\overrightarrow{ED}=\left( pc,pd \right)$

Moreover, $G\text{ is the midpoint of }AC\Leftrightarrow G\left( \dfrac{a+c}{2},\dfrac{d}{2} \right)$ $H\text{ is the midpoint of }BD\Leftrightarrow H\left( \dfrac{b+pc}{2},\dfrac{pd}{2} \right)$ Now we proceed to the calculation of the areas. $\begin{aligned} \left[ ABCD \right] & =\left[ EAD \right]-\left[ EBC \right] \\ & =\frac{1}{2}\left| \det \left( \overrightarrow{EA},\overrightarrow{ED} \right) \right|-\frac{1}{2}\left| \det \left( \overrightarrow{EB},\overrightarrow{EC} \right) \right| \\ & =\frac{1}{2}\left| \left| \begin{matrix} a & 0 \\ pc & pd \\ \end{matrix} \right| \right|-\frac{1}{2}\left| \left| \begin{matrix} b & 0 \\ c & d \\ \end{matrix} \right| \right| \\ & =\frac{1}{2}\left| apd \right|-\frac{1}{2}\left| bd \right| \\ \end{aligned}$ We notice that $a$ , $b$ , $p$ , $d$ are all positive numbers, thus $\left[ ABCD \right]=\frac{1}{2}\left( apd-bd \right) \ \ \ \ \ (1)$ For $\triangle EHG$ , $\begin{aligned} & \left[ EHG \right]=\dfrac{1}{2}\left| \det \left( \overrightarrow{EG},\overrightarrow{EH} \right) \right| \\ & =\dfrac{1}{2}\left| \left| \begin{matrix} \dfrac{a+c}{2} & \dfrac{d}{2} \\ \dfrac{b+pc}{2} & \dfrac{pd}{2} \\ \end{matrix} \right| \right| \\ & =\dfrac{1}{8}\left| apd-bd \right| \\ \end{aligned}$ But, $a>b>0$ , $p>1$ and $d>0$ imply that $apd>bd$ , thus, $\left[ EHG \right]=\dfrac{1}{8}\left( apd-bd \right) \ \ \ \ \ (2)$ Finally, $\left( 1 \right),\left( 2 \right)\Rightarrow \dfrac{\left[ EHG \right]}{\left[ ABCD \right]}=\dfrac{\dfrac{1}{8}\left( apd-bd \right)}{\dfrac{1}{2}\left( apd-bd \right)}=\dfrac{1}{4}=\boxed{0.25}$ .