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