A linear algebra solution to a euclidean geometry problem While i was studying linear algebra i came up with this problem:
Let $ \triangle ABC$ be a triangle and $P,Q$ be two points in the plane such that segments $PQ$ and $AC$ share the midpoints and both $P,Q$ are on the same side of line $AB$. Prove:
$$ [ABC]=[APB]+[AQB] $$ 
Where $[PQR]$ is the area of triangle $PQR$.
The "linear algebra" proof i found is:
Since $AC$ and $ PQ $ share midpoints, then considering vectors $ \displaystyle \frac{\vec{A}+\vec{C}}{2}=\frac{\vec{P}+\vec{Q}}{2} $ and setting $A$ as the origin we get $\vec{C}=\vec{P}+\vec{Q}$. 
Now we use areas as determinants, let $C=(x_C,y_C) $ in the plane and other points's coordinates defined in the same way we have:
$$ [ABC]=\frac{1}{2} \left| \begin{array}{c} x_C & x_B\\ y_C & y_B  \end{array} \right|=\frac{1}{2} \left| \begin{array}{c} x_P+x_Q & x_B\\ y_P+y_Q & y_B  \end{array} \right|=\frac{1}{2} \left| \begin{array}{c} x_P & x_B\\ y_P & y_B  \end{array} \right|+  \frac{1}{2} \left| \begin{array}{c} x_Q & x_B\\ y_Q & y_B  \end{array} \right|=[APB]+[AQB]$$
and this is our desired identity.
My question is: is there a full geometric motivation to this fact? Can we give a proof that does not involve coordinates or trig bash?
 A: 
Let $M$ be the midpoint of $AC$. Since $PQ$ and $AC$ share the same midpoint we have that $P$ and $Q$ are symmetric with respect to $M$. We have $AM=MC$ and the triangles $AMB,CMB$ share the altitude through $B$, hence $[AMB]=[CMB]=\frac{1}{2}[ABC]$. In a similar way we have $[BPM]=[BMQ]$ and $[PAM]=[MAQ]$, hence:
$$ [APB]+[AQB] = 2\left([APB]+[PAM]+[PBM]\right) = 2 [ABM] = [ABC] $$
as wanted. I leave to you to check there are no configuration issues, since we may assume that $AB$ lies on the $x$-axis, $C$ lies in the upper half-plane, the $y$-coordinate of $Q$ is $\geq$ than the $y$-coordinate of $P$, with both $y$-coordinates being non-negative.
A: Let $D$ be the common midpoint of $AC$ and $PQ\,$, then using signed areas:
$$
[APB] = [DBA]+[DAQ]+[DQB] \\
[AQB] = [DBA]+[DAP]+[DPB]
$$
Adding the two:
$$
[APB]+[AQB] = 2 [DBA]+[DAQ]+[DQB] +[DAP]+[DPB]
$$
Note that:


*

*$2 [DBA] = [ACB]$ since $[DBA]=[DCA]$ because $\triangle BAD$ and $\triangle BDC$ have the same height and equal bases $AD=DC\,$;

*$[DAQ] + [DAP] = 0$ since $\triangle AQD$ and $\triangle ADP$ have opposite orientation, but the same height and equal bases $QD=DP\,$;

*$[DQB] + [DPB] = 0$ by a similar argument.
Therefore $\require{cancel}[APB]+[AQB] = [ACB]+\cancel{[DAQ]}+\bcancel{[DQB]} +\cancel{[DAP]}+\bcancel{[DPB]} = [ACB]\,$.
A: Here is a coordinates' free proof, denoting by $M$ the common point:
$$[ABC]=\tfrac12 \det(\vec{AB},\vec{AC})=\tfrac12 \det(\vec{AB},2 \vec{AM})=$$
$$\tfrac12 \det(\vec{AB},\vec{AP}+\vec{AQ})=\tfrac12 \det(\vec{AB},\vec{AP})+\tfrac12 \det(\vec{AB},\vec{AQ}),$$
proving that :
$$[ABC]=[APB]+[AQB]$$
