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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

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    $\begingroup$ @dfnu yes sorry $\endgroup$
    – Tyrone
    Commented Nov 19, 2019 at 13:22

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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 $A+B=D+E=\pi$?)

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  • $\begingroup$ How did you get the $\frac{1}{2}AF.BC.sin(A+B)$. Wouldn’t that only work if $AB$ was parallel to $FC$? $\endgroup$
    – Tyrone
    Commented Nov 20, 2019 at 12:35
  • $\begingroup$ If you work out the areas of the three triangles in terms of the lengths and angles, AK,AF,A etc then you'll see how it all drops out. I will return to the website later today if you have any further questions. $\endgroup$
    – user502266
    Commented Nov 20, 2019 at 12:50
  • $\begingroup$ I've written out the formulae now. There was a typo (a wrong sign) which may have worried you - sorry if that was the case. $\endgroup$
    – user502266
    Commented Nov 20, 2019 at 18:20

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