A question on proving a midpoint when ratio of the area of triangles is given In a convex quadrilateral $PQRS$, the areas of triangles $PQS$, $QRS$ and $PQR$ are in the ratio $3 : 4 : 1$. A line through $Q$ cuts $PR$ at $A$ and $RS$ at $B$ such that $PA : PR = RB : RS$. Prove that $A$ is the midpoint of $PR$ and $B$ is the midpoint of $RS$.
This question appeared in a competitive mathematics exam. What I am doing is taking the ratios of the areas of triangles given and then equating this ratio to the respective sides of those triangles, like for example,
$PA : PR = RB : RS, \Rightarrow [PQA]:[PQR]=[QRB]:[QRS]$
Then, I am assuming that one of the things to be proved, $A$ is the midpoint of $PR$ (which I have proved implies that $B$ is the midpoint of $RS$), to be correct, and using this result, showing that these conditions comply with the assumption. But I am stuck at how to prove that these results comply for these assumption exclusively.
Please help. Thanks!
 A: 
As we know the ratios of the areas of internal triangles, we have
$PO:OR = 3:4, QO:OS = 1:6$
We assume $OA = y, PO = 3x$.
Now applying Menelaus's theorem in $\triangle SOR$ where line $BQ$ cuts the triangle.
$\displaystyle \frac{SQ}{QO} \cdot  \frac{OA}{AR} \cdot \frac{RB}{BS} = 1$
$\displaystyle \frac{7y}{4x-y} = \frac{RS}{RB}-1 = \frac{PR}{PA}-1$
$\displaystyle \frac{7y}{4x-y} = \frac{7x}{3x+y}-1 = \frac{4x-y}{3x+y}$
$6y^2+29xy-16x^2 = 0 \implies (3y+16x) (2y-x) = 0$
That leads to $\displaystyle y = \frac{x}{2}$ and $\displaystyle PA = AR = \frac{7x}{2}$. As $A$ is midpoint of $PR$, so is $B$ the midpoint of $RS$.
A: Here is a solution using barycentric coordinates which is a way to work on ratios of areas. They are often useful in Mathematical Contests.
See the following picture.

Fig. 1. a,b,c,d are areas delimited by diagonals.
Without loss of generality, we can assume that the area of $PQRS$ is $1.$
Diagonals $PR$ and $QS$ define four (unsigned) areas denoted $a,b,c,d$.
The given ratios $3:4:1$ give the following relationships:
$$\begin{cases}a+b&=&3/7&(a)\\c+d&=&4/7&(b)\\b+c&=&1/7&(c)\end{cases}\tag{1}$$
Remark: combining (a)+(b)-(c), one gets:
$$a+d=6/7\tag{2}$$
Let us consider $PRS$ as the basis triangle, with barycentric coordinates of generic point $M$ given by quotients of oriented areas:
$$M=\begin{cases}p=area(MRS)/area(PRS)\\r=area(PMS)/area(PRS)\\  s=area(PRM)/area(PRS)\end{cases} $$
Using (1) and (2), it is easy to obtain the barycentric coordinates of :
$$Q=\begin{cases}
p_Q&=& \ \ \ (c+d)/(a+d)&=& \ \ \ 4/6\\
r_Q&=& \ \ \ (a+b)/(a+d)&=& \ \ \ 3/6\\
s_Q&=&-(b+c)/(a+d)&=&-1/6\end{cases}\tag{3}$$
Please note that the minus sign for $s_Q$ comes from the fact that oriented area$(PRQ)<0$.
Besides, if  $A$ and $B$ are defined as the midpoints of $PR$ and $SR$ resp., their barycentric coordinates are
$$A=\begin{cases}
p_A&=&1/2\\
r_A&=&1/2\\
s_A&=&0\end{cases}, \ \ \ \ \ 
B=\begin{cases}
p_B&=&0\\
r_B&=&1/2\\
s_B&=&1/2\end{cases}$$
As the determinant
$$\begin{vmatrix}p_A&p_B&p_Q\\
r_A&r_B&r_Q\\
s_A&s_B&s_Q
\end{vmatrix}=0$$
we conclude that $A,B,Q$ are aligned.
Edit: An even shorter way would have been, after (3), to use the fact that $r_Q=1/2$ to conclude that $Q$ belongs to the midline $AB$ whose barycentric equation is clearly $r=1/2$.
