In $\triangle ABC$, $D$ is the mid point of $AB$.. In $\triangle ABC$, $D$ is the mid point of $AB$ and $P$ is any point on $BC$. If $CQ||DP$ then prove that: $2\triangle BPQ=\triangle ABC$.
My Attempt ;
Since $D$ is the mid point of $AB$ and $DP||QC$ then $P$ is the mid point of $BC$ I.e  by mid point Theorem 
 A: Construction:-$CD$ is joined.
Proof:-
$\triangle DPQ=\triangle DCP$(same base and same parallels)
Adding $\triangle BPD$ on both sides we get,
$\triangle DPQ+\triangle BPD=\triangle DCP+\triangle BPD$
$\implies \triangle BPQ=\triangle BDC$
Now,$2\triangle BDC=\triangle ABC$(median divides triangle into two equal areas)
$\implies2\triangle BPQ=\triangle ABC$
Hence,proved.
By the way, $D$ is the mp of $AB$..not $BQ$...hence your mid-point theorem statement is wrong....(Be careful of which triangle you are applying mid point theorem on)...
A: Hint:  $[DQP]=[DCP]$ 
And $2[DBC]=[ABC]$  
A: We observe that
$$
\triangle BQP = \color{blue}{\triangle BDP} + \color{red}{\triangle DPQ},\quad \triangle ABC = \color{blue}{\triangle ABP}+\color{red}{\triangle APC}
$$
so it's enough to show
$$
\color{blue}{\triangle BDP}=\frac{1}{2}\color{blue}{\triangle ABP}\quad\text{and}\quad \color{red}{\triangle DPQ}=\frac{1}{2}\color{red}{\triangle APC}.
$$
The first equality above follows because $D$ is the midpoint of $AB$. The second uses the parallel condition and then again the fact that $D$ is the midpoint of $AB$:
$$
\triangle DPQ=\triangle DPC=\frac{1}{2}\triangle APC.
$$
Note that $\triangle DPC$ and $\triangle APC$ shares the same base $PC$ and the former triangle has altitude that is half of that of the latter.
