# If $AB\parallel DC$, $BC\parallel AD$, and $AC\parallel DQ$, find $\Bbb X$ in terms of the areas $\Bbb A$, $\Bbb B$, and $\Bbb C$.

If $$AB\parallel DC$$, $$BC\parallel AD$$, and $$AC\parallel DQ$$, find $$\Bbb X$$ in terms of the areas $$\Bbb A$$, $$\Bbb B$$, and $$\Bbb C$$.

Please, I wrote a lot of relations, but I just need to prove that $$\overline{AC}\cap\overline{BQ}=P\implies BP=PQ$$. If I prove this, I will get $$\Bbb X=\Bbb A+2\Bbb B+\Bbb C.$$

For context, if $$BP=PQ$$, then $$\triangle BPC$$ and $$\triangle CPQ$$ have equal area. Let $$\overline{BP}\cap\overline{CD}=R$$. Then, we can show that $$\triangle APR$$ and $$\triangle BRC$$ have equal area. If $$S=\overline{AQ}\cap\overline{CD}$$, then we can see that $$\mathbb{X}-\mathbb{A}-\mathbb{B}$$ is the area of $$\triangle ARS$$. It is not difficult to show that the area of $$\triangle ARS$$ is $$\mathbb{B}+\mathbb{C}$$.

• And I couldn't proof this. – Tas Apr 28 '20 at 11:36
• You have a parallelogram $ABCD$ and a diagonal $AC$. Then you have a parallel passing by $D$ but you say nothing about the position of the point $Q$. Edit your problem. – Piquito Apr 28 '20 at 12:58
• @Piquito That is because the location of $Q$ is irrelevant, as long as $CD$ intersects $BQ$ and $AQ$ internally. The equality $\Bbb X=\Bbb A+2\Bbb B+\Bbb C$ holds regardless. – Batominovski Apr 28 '20 at 13:06

Here is a solution to the problem. Let $$[S]$$ denote the area of a shape $$S$$. Define $$P=AC\cap BQ$$, $$M=BQ\cap CD$$, and $$N=AQ\cap CD$$. The line $$DQ$$ meets the line $$AB$$ and $$BC$$ at $$E$$ and $$F$$, resp. The line passing through $$M$$ parallel to $$AC\parallel DQ$$ meets $$AN$$ at $$R$$.

Since $$QA$$ is a median of $$\triangle QBE$$ and $$DM\parallel MD$$, $$QN$$ is also a median of $$\triangle QMD$$. Therefore $$\Bbb C=[DNQ]=[MNQ]$$.

Thus $$\Bbb B+\Bbb C=[MNQ]+[CMQ]=[CNQ]$$. Because $$MN=ND$$ and $$RM\parallel DQ$$, we easily see that $$RN=NQ$$. Thus $$[CNR]=[CNQ]=\Bbb B+\Bbb C$$. This shows that \begin{align}[ARC]&=[ANC]-[CNR]\\&=[APMN]+[PMC]-[CNR]\\&=\Bbb X+\Bbb A-\Bbb B-\Bbb C.\end{align}

Because $$RM\parallel AC$$, we get $$[ARC]=[AMC]$$. As $$AB\parallel CD$$, we have $$[AMC]=[BMC]=[BPC]+[PMC].$$ Since $$P$$ is the midpoint of $$BQ$$ as proven by timon92, we have $$[BPC]=[QPC]=[PMC]+[CMQ]=\Bbb A+\Bbb B.$$ Thus $$[AMC]=[BPC]+[PMC]=(\Bbb A+\Bbb B)+\Bbb A=2\Bbb A+\Bbb B.$$ Thus, $$\Bbb X+\Bbb A-\Bbb B-\Bbb C=[AMC]=2\Bbb A+\Bbb B,$$ or $$\Bbb X=\Bbb A+2\Bbb B+\Bbb C.$$

Here is how to use coordinate geometry to solve this problem. WLOG, we can assume that $$ABCD$$ is a unit square: $$A=(0,0)$$, $$B=(0,1)$$, $$C=(1,1)$$, and $$D=(1,0)$$. Let $$Q=(1+t,t)$$ where $$0\le t\le 1$$.

Then $$AQ$$ is given by $$y=\frac{t}{1+t}x.$$ So $$N=\left(1,\frac{t}{1+t}\right)$$. Therefore $$\Bbb C=[DNQ]=\frac12 \cdot\frac{t}{1+t}\cdot t=\frac{t^2}{2(1+t)}.$$

The line $$BQ$$ is given by $$y-1=-\frac{1-t}{1+t}x.$$ Therefore $$M=\left(1,\frac{2t}{1+t}\right).$$ Hence $$\Bbb B=[CMQ]=\frac12\cdot \left(1-\frac{2t}{1+t}\right)\cdot t=\frac{t(1-t)}{2(1+t)}.$$

We also have $$P=\left(\frac{1+t}{2},\frac{1+t}{2}\right).$$ Hence $$\Bbb A=[CPM]=\frac12\cdot\left(1-\frac{2t}{1+t}\right)\cdot\left(1-\frac{1+t}{2}\right)=\frac{(1-t)^2}{4(1+t)}.$$ Now $$\Bbb X+\Bbb A=[ANC]=\frac12\cdot\left(1-\frac{t}{1+t}\right)\cdot 1=\frac{1}{2(1+t)}.$$ Hence $$\Bbb X=\frac{1}{2(1+t)}-\frac{(1-t)^2}{4(1+t)}=\frac{1+2t-t^2}{4(1+t)}.$$ If $$\Bbb X=a\Bbb A+b\Bbb b+c\Bbb C$$, then $$1+2t-t^2=a(1-t)^2+2bt(1-t)+2ct^2=a+(-2a+2b)t+(a-2b+2c)t^2.$$ That is, $$a=1$$, $$b=2$$, and $$c=1$$.

Well, $$ABCD$$ is a parallelogram so $$AC$$ bisects $$BD$$. Using this and $$AC\parallel DQ$$ we obtain $$BP=PQ$$.

• Why? The position of point $Q$ is not fixed. – Piquito Apr 28 '20 at 13:01
• @Piquito Just look at the triangle $BDQ$. Line $AC$ is parallel to $DQ$ and bisects $BD$. Hence it also bisects $BQ$. – timon92 Apr 28 '20 at 13:03