Notes
subdivided triangle solution

Subdivided Triangle

Solution by Pythagoras' Theorem and Crossed Trapezium

Since the yellow segments have length $3$ and the red have length $5$, the side lengths of the triangle are $6$, $8$, and $10$. These satisfy $10^2 = 6^2 + 8^2$ which means that triangle $A D C$ is right-angled by the converse to Pythagoras' Theorem.

As $E$ is the midpoint of $A D$ and $B$ of $A C$, $E B$ is parallel to $D C$ and so $E B C D$ is a trapezium. With the diagonals, this makes a crossed trapezium. The scale factor from $E B$ to $D C$ is $2$, so the area of the trapezium is $(1 + 2)^2 = 9$ times the area of triangle $E G B$. Since $E B$ has length $4$, $D C$ length $8$, and $E D$ has length $3$ the area of the trapezium is $18$ so triangle $E G B$ has area $2$. Then triangle $E G D$ has area $4$.