Relation between areas of two non-similar triangles with individual proportional sides Triangle $ABC$ has area $k$ and $D$ is the middle point of $BC$.
We have $AP = 2 \cdot AB$, $AQ = 3 \cdot AD$ and $AR = 4 \cdot AC$. What's the area of triangle $PQR$?
I know that the answer is $k$, but I don't know how to prove it. 
Thank you all in advance!



Let $\angle PAQ = \alpha$ and $\angle RAQ = \beta$.
$$A_{PQR} = A_{APQ} + A_{AQR}- A_{APR}$$
$$=\frac12 AP\cdot AQ\sin\alpha + \frac12 AR\cdot AQ\sin\beta - \frac12 AP\cdot AR\sin(\alpha+\beta)$$
$$=\frac12 ( 2AB \cdot 3AD\sin\alpha )+ \frac12 ( 3AD \cdot 4AC\sin\beta)
- \frac12 ( 2AB \cdot4AC\sin(\alpha+\beta))$$
$$= 6 A_{ABD}+ 12 A_{ACD} - 8A_{ABC}
=6\cdot\frac k2+12\cdot\frac k2-8k=k$$
