Let $ABCD$ be a [parallelogram](https://artofproblemsolving.com/wiki/index.php/Parallelogram).  Extend $\overline{DA}$ through $A$ to a point $P,$ and let $\overline{PC}$ meet $\overline{AB}$ at $Q$ and $\overline{DB}$ at $R.$  Given that $PQ = 735$ and $QR = 112,$ find $RC.$
Solution 1
[AIME 1998-6.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_1998-6.png)
There are several [similar triangles](https://artofproblemsolving.com/wiki/index.php/Similar_triangles). $\triangle PAQ\sim \triangle PDC$, so we can write the [proportion](https://artofproblemsolving.com/wiki/index.php/Proportion):


$\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}$


Also, $\triangle BRQ\sim DRC$, so:


$\frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD}$
$\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}$


Substituting,


$\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}$
$735RC = (RC + 847)(RC - 112)$
$0 = RC^2 - 112\cdot847$


Thus, $RC = \sqrt{112*847} = 308$.

Solution 2
We have $\triangle BRQ\sim \triangle DRC$ so $\frac{112}{RC} = \frac{BR}{DR}$. We also have $\triangle BRC \sim \triangle DRP$ so $\frac{ RC}{847} = \frac {BR}{DR}$. Equating the two results gives $\frac{112}{RC} =  \frac{ RC}{847}$ and so $RC^2=112*847$ which solves to $RC=\boxed{308}$