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# The parallelogram ABCD shows the points P and Q dividing each of the line AD and DC in the ratio 1:4. What is the ratio in which R divides DB? What is the ratio in which R divides PQ?

Last updated date: 16th Sep 2024
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Hint: First construct the line AQ parallel to PQ. Using similar triangles DPQ and DAQ. There are many types of theorems like cevian theorem to get the ratio of the DB and PQ which pass through R.

Let $\left| {ABC} \right|$ denote area of $\vartriangle ABC$
Let $\left| {ABCD} \right| = 10x$,
$\Rightarrow \left| {ABC} \right| = \left| {ADC} \right| = \left| {ABD} \right| = \left| {CBD} \right| = \dfrac{{10x}}{2} = 5x$
Given, $DQ:QC = 1:4$,
$\Rightarrow \left| {ADQ} \right|:\left| {AQC} \right| = \left| {BDQ} \right|\left| {BQC} \right| = 1:4$,
$\Rightarrow \left| {ABP} \right| = x,\left| {PBD} \right| = 4x$,
And, $\left| {AQP} \right| = \dfrac{1}{5}x,\left| {PQD} \right| = \dfrac{4}{5}x$
$\left| {PDQB} \right| = \left| {PBD} \right| + \left| {BDQ} \right| = 4x + x = 5x$
$\Rightarrow \left| {PQB} \right| = \left| {PDQB} \right| - \left| {PDQ} \right| = 5x - \dfrac{4}{5}x = \dfrac{{21}}{5}x$
$\Rightarrow DR:RB = \left| {PDQ} \right|:\left| {PQB} \right| = \dfrac{4}{5}x:\dfrac{{21}}{5}x = 4:21$
$\therefore PR:RQ = \left| {BPD} \right|:\left| {BDQ} \right| = 4x:x = 4:1$