Question

P is mid point of the side $CD$ of a parallelogram ABCD. A line through $C$ parallel to $PA$ intersect $AB$ at $Q$ and $DA$ produced at $R.$ Prove that $DA=AR$ and $CQ=QR.$

Answer 1

Refer attached image, given ABCD is a parallelogram

$AP\parallel CR$

$CP=PD$

Since, $△APD\sim △RCD\left(AAA\right)$

$\Rightarrow \frac{DA}{DR}=\frac{DP}{DC}\left(CPCT\right)$

$\Rightarrow \frac{DA}{DR}=\frac{1}{2}\left(AsPisthemidpointofCD\right)$

$\Rightarrow 2DA=DR$

$\Rightarrow 2DA=DA+AR$

$\Rightarrow DA=AR$

Hence Proved DA=AR

Again as, $\angle APD=\angle QCD\&\angle BQC=\angle APD(AP\parallel QC)$,

$\Rightarrow \angle APD=\angle QCD=\angle BQC$

&

$\angle RDC=\angle CBQ(Oppositeof\angle is\parallel gm)$

&

$\angle BCQ=\angle CRD(Propertiesof\parallel gm)$

$\Rightarrow △BQC\sim △DCR\left(AAA\right)$

Now,

As,

$\Rightarrow △APD\cong △CBQ$

$PD=BQ$

or $\frac{CD}{2}=BQ\left(AsPisthemidpointofCD\right)$

or $\frac{AB}{2}=BQ(In\parallel gmAB=CD)$

$\Rightarrow$ Q is the mid-point of AB

Again,

as,

$△BQC\sim △DCR$

$\frac{BQ}{CQ}=\frac{CD}{CR}$

$\Rightarrow 2CQ=CR(AsCD=2BQ)$

$\Rightarrow 2CQ=CQ+QR$

$\Rightarrow CQ=QR$

Hence Proved CQ=QR