BdMO - 2018 Regional - Geometry 9 In $ABCD$ tetragon $E $ and $F $ are mid points of $AB$ and $AD$ respectively. $CF$
intersects $BD$ at point $G$. If $\angle FGD = \angle AEF$ and the area of $ABCD$ is $24$ , what is
the area of $ABCG$ ?


Note that $\angle FGD=\angle AEF$ and $EF\parallel BD$ so it must be that $FG 
\parallel AB$. So triangles $FGD$ and $AEF$ have equal corresponding angles. On the other side $AF=FD$ and by ASA triangles $FGD$ and $AEF$ are congruent. It follows immediatelly that $EBGF$ is a parallelogram.
$$A_{ABCD}=A_{ABD}+A_{BCD}=\frac{2a\cdot 2h}{2}+2\cdot\frac{bh}2=(2a+b)h\tag{1}$$
In a similar way for trapezoid $ABCG$:
$$A_{ABCG}=\frac{2a+b}{2}h\tag{2}$$
By comparing (1) and (2) it's obvious that:
$$A_{ABCG}=\frac{1}{2}A_{ABCD}=12$$
