Question

In the figure above, a circle passing through vertex $A$ of the parallelogram intersects the sides and the diagonal as shown. Which of the following is/are correct:

• A. $AQ\times AC=AP\times AB+AR\times AD$
• B. $\frac{AQ}{AC}$ $=$ $\frac{AP}{AB}$ $+$ $\frac{AR}{AD}$
• C. $\frac{QC}{PB}$ $=$ $\frac{PB}{DR}$
• D. None of the Above

Answer 1

The correct option is A $AQ\times AC=AP\times AB+AR\times AD$

Applying Ptolemy's Theorem
to cyclic quadrilateral $APQR$ we have:
$PR\ast AQ=PQ\ast AR+RQ\ast AP$ $\left(1\right)$
Now we'll prove that $\Delta PQR$ and $\Delta ADC$ are similar. Since $APQR$ is cyclic we have:
$\angle RPQ=\angle RAP=\angle DAC$

$\angle PRQ=\angle QAP=\angle CAB$

but $AB||CD$ and $\angle CAB=\angle DCA$, Thus $\Delta ADC\sim \Delta PQR$ as a consequence:
$\frac{RQ}{CD}$ = $\frac{QP}{DA}$ =$\frac{PR}{AC}$ $\left(2\right)$

but since $ABCD$ is parallelogram $CD=AB$ thus the identities
in $\left(2\right)$ can be written as:
$PR=AC\times$ $\frac{RQ}{AB}$

$QP=DA\times$ $\frac{RQ}{AB}$

Now by putting these in $\left(1\right)$ we have:

$AQ\ast AC=AP\ast AB+AR\ast AD$