Prove that the sum of the areas of triangles $FAK$, $KCB$, and $CFL$ is equal to half of the hexagon.

Question

Let $ABCDEF$ be a convex hexagon with $\angle A=\angle D$ and $\angle B=\angle E$. Let $K$ and $L$ be the midpoints of the sides $AB$ and $DE$ respectively.

Prove that the sum of the areas of triangles $FAK$, $KCB$, and $CFL$ is equal to half of the area of the hexagon if and only if $$\frac{BC}{CD}=\frac{EF}{FA}$$

I cant figure out how to relate the proportions of the sides to the area, suggestions as-well as solutions would be appreciated

Taken from the 2013 Pan African Maths Olympiad http://pamo-official.org/problemes/PAMO_2013_Problems_En.pdf

Answer 1

Consider quadrilateral $ABCF$. There is an elegant formula for the areas:-

$$\text {Area }CFK=\text {Area }AKF+\text {Area }BCK- \frac{1}{2}AF.BC.\text{sin}(A+B).$$

Proof

Drop perpendiculars from $C$ and $F$ onto the line $AB$ and let $AK=KB=K$. Then

$2\times\text {Area }AKF=kAF\text{sin}A,\,\,\,\,2\times\text {Area }BKC=kBC\text{sin}B$.

$2\times\text {Area }FKC=(2k-AF\text{cos}A-BC\text{cos}B)(AF\text{sin}A+BC\text{sin}B)$

$$-(k-AF\text{cos}A)AF\text{sin}A-(k-BC\text{cos}B)BC\text{sin}B.$$ Multiplying out now gives the required result.

Similarly, for quadrilateral $DEFC$:-

$$\text {Area }FCL=\text {Area }CDL+\text {Area }EFL- \frac{1}{2}DC.EF.\text{sin}(D+E).$$

For equal areas we require $$\text {Area }FCL+\text {Area }AKF+\text {Area }BCK=\text {Area }CFK+\text {Area }CDL+\text {Area }EFL$$ and therefore

$$\frac{1}{2}AF.BC.\text{sin}(A+B)=\frac{1}{2}DC.EF.\text{sin}(D+E).$$ i.e. if and only if $AF.BC=DC.EF$. (Or $A+B=D+E=\pi$?)