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$2.$ In the convex quadrilateral $ABCD$, points $M,N$ lie on the side $AB$ such that $AM = MN = NB$, and points $P,Q$ lie on the side $CD$ such that $CP = PQ = QD$. Prove that Area of $AMCP=$ Area of $MNPQ = \frac{1} {3} $ Area of $ABCD$.

I know how to prove that $AMCP=\frac{1}{3}$ Area of $ABCD$, but not sure about the rest. Would it suffice to say that the other section is just a third of a square that has been stretched or does this need to be proven? Thanks in advance to anyone who can prove or come up with any hints.

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It will suffice to prove that distance from $M$ to $CD$ $\rho(M,CD)=\frac{2}{3}\rho(A,CD)+\frac{1}{3}\rho(B,CD)$ and $\rho(N,CD)=\frac{1}{3}\rho(A,CD)+\frac{2}{3}\rho(B,CD)$ together with $\rho(P,AB)=\frac{2}{3}\rho(C,AB)+\frac{1}{3}\rho(D,AB)$ and $\rho(Q,AB)=\frac{1}{3}\rho(C,AB)+\frac{2}{3}\rho(D,AB)$, then make a triangulation.
I'll do about $NMPQ$: Area $S_{MNPQ}=S_{PQN}+S_{QMN}=\frac{1}{2}\left(PQ\cdot\rho(N,CD)+MN\cdot\rho(Q,AB)\right)= \frac{1}{2}\left(\frac{1}{3}CD\cdot\left(\frac{1}{3}\rho(A,CD)+\frac{2}{3}\rho(B,CD)\right)+\frac{1}{3}AB\cdot\left(\frac{1}{3}\rho(C,AB)+\frac{2}{3}\rho(D,AB)\right)\right)= \frac{1}{18}\left(CD\cdot \rho(A,CD)+2CD\cdot \rho(B,CD)+AB\cdot \rho(C,AB)+2AB\cdot \rho(D,AB) \right)= \frac{1}{9}\left(S_{ACD}+2S_{BCD}+S_{ABC}+2S_{ABD}\right)=\frac{1}{3}S_{ABCD}$
Now we want to prove about the distances. Wlog $D$ is closer to $AB$ than $C$ is (in the case $\rho(D,AB)=\rho(C,AB)$ $AB||CD$, and $\rho(X,AB)=const$ for every point $X$ on $CD$).
Let $DC_1||AB$ and the perpendiculars from $Q,P,C$ to $AB$ intersects $DC_1$ in the points $Q_1,P_1,C_1$ resp. $QQ_1||PP_1||CC_1\Rightarrow \Delta DQQ_1,\Delta DPP_1, \Delta DCC_1$ are similar $\Rightarrow QQ_1=\frac{1}{3}CC_1, PP_1=\frac{2}{3}CC_1$ hence we got the demanded equalities for distances.

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  • $\begingroup$ What exactly do you mean by the distance rho, is this the shortest distance between the two points? $\endgroup$ – MadChickenMan May 20 '15 at 21:39
  • $\begingroup$ I mean the distance between two geometrical objects. In this case between a point and a line, i.e. length of the perpendicular, dropped from point to line. $\endgroup$ – Alexey Burdin May 20 '15 at 21:57
  • $\begingroup$ Ah. I have actually just managed to come up with a different proof to the question, but I'm still interested in the first part of your solution. I'm not really sure how to prove the first statement. $\endgroup$ – MadChickenMan May 20 '15 at 22:03
  • $\begingroup$ Edited, now you can see the full proof. :) $\endgroup$ – Alexey Burdin May 20 '15 at 22:39
  • $\begingroup$ I see now. Thanks for all your help. $\endgroup$ – MadChickenMan May 21 '15 at 6:48
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I would use vector notations (based on picture above). Use $\times$ as cross product. To make sure that all areas sum with correct sign - follow clockwise rotation. Then: $$S_{ADCB}*2 = |\vec{CA}\times\vec{CD} + \vec{AC}\times\vec{AB}| = |\vec{CA}\times\vec{CP*3} + \vec{AC}\times\vec{AM*3}|$$ and $$S_{ADCB} = 3 * S_{AMCP}$$ $$S_{MNPQ}*2 = |\vec{PM}\times\vec{PQ} + \vec{MP}\times\vec{MN}| = |(\vec{PC}+\vec{CA}+\vec{AM})\times\vec{CP} + (\vec{MA}+\vec{AC}+\vec{CP})\times\vec{AM}|$$ Now remember that cross product of parallel vectors is zero and cross product is anticommutative. $$S_{MNPQ}*2 = |(\vec{CA}\times\vec{CP}+\vec{AM}\times\vec{CP} + \vec{AC}\times\vec{AM}-\vec{AM}\times\vec{CP}|$$ $$S_{MNPQ} = S_{AMCP}$$

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  • $\begingroup$ Thanks for your answer. Afraid I'm not well versed in vector mathematics yet so I'm not able to full appreciate your proof though. $\endgroup$ – MadChickenMan May 21 '15 at 6:49

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