BdMO - 2018 Regional - Geometry 9

Question

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 $ABCG$ ?

Answer 1

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:

$$A_{ABCG}=\frac{1}{2}A_{ABCD}=12$$

Answer 2

Since $EF||BD$ ($EF$ is a midlle line in $ABD$) and $\angle FGD = \angle AEF$ we have $EG||AB$ so $G$ halves $BD$. If we use now the fact that median halves triangle on two triangles with equal area, we have $$ABCG = ABG+BCG= {1\over 2}ABD+{1\over 2}BCD = {1\over 2}ABCD = 12$$