As shown in the figure below, the area of rectangle $ABCD = 1$, $Q$ and $P$ are points on $CD$ and $BC$.
Such that $\triangle ADQ = \triangle QCP = \triangle ABP$
Find the area of $\triangle ADQ$.
I've tried assuming the three triangles' area is $a$ each, $AD=x, AB=y$, then figuring out $DQ, BP$ by areas of $\triangle ADQ, \triangle ABP$, then making an equation for the area of $\triangle QPC$.
Only knowing $xy=1$, my method doesn't seem to work very well. I get $x^2+y^2$ terms at the end. Is there a way to find $x^2+y^2$?
Best Answer
Following the OP's notation for areas and lengths, we find $DQ=\dfrac{2a}x$ and $BP=\dfrac{2a}y$.
Therefore $PC=x-\dfrac{2a}y$ and $QC=y-\dfrac{2a}x.$
Now considering the area of $\triangle PCQ$,
$$a=\frac12\left(x-\frac{2a}y\right)\left(y-\frac{2a}x\right)$$ $$2a=xy-4a+\frac{4a^2}{xy}$$ $$4a^2-6a+1=0$$
(because $xy=1$)
And $a$ is the area required. Can you take it?