BMO1 2006/07 Question 2 Geometry Problem $2.$ 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.
 A: 
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. 
$$S_{MNPQ}*2 = |(\vec{CA}\times\vec{CP}+\vec{AM}\times\vec{CP} + \vec{AC}\times\vec{AM}-\vec{AM}\times\vec{CP}|$$
$$S_{MNPQ} = S_{AMCP}$$
A: 
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.
