ABCD is a parallelogram. AB is divided at P and CD at Q such that AP : PB is 3 : 2 and CQ : QD is 4 : 1. if PQ meets AC at R, prove that AR= 3/7 AC ?

GIVEN : ABCD is a ||^gm , in which P is a point on AB such that AP : PB = 3 : 2 and Q on CD such that CQ : QD = 4 : 1.
TO PROVE : $AR = \frac{3}{7}AC$

PROOF : Since ABCD is a ||^gm , then AB || CD and AD || BC.
Since AB || CD and AC is a transversal, so
$\angle PAR = \angle QCR \left(Alternate interior angles\right) .........\left(1\right)$

Since ABCD is a ||^gm , then AB = CD and AD = BC , as opposite sides of ||^gm are equal.
Let AB = CD = x
$$\begin{gathered} \mathrm{so}, \mathrm{AP} = \frac{3x}{5} \mathrm{and} \mathrm{PB} = \frac{2x}{5} \left[\mathrm{as}, \mathrm{AP}:\mathrm{PB} = 3:2\right] \\ \mathrm{and} \mathrm{CQ} = \frac{4x}{5} \mathrm{and} \mathrm{QD} = \frac{x}{5} \left[\mathrm{as}, \mathrm{CQ}:\mathrm{QD} = 4:1\right] \end{gathered}$$

$$\begin{gathered} \mathrm{In} △\mathrm{CQR} \mathrm{and} △\mathrm{APR}, \\ \angle \mathrm{QCR} = \angle \mathrm{PAR} \left[\mathrm{using} \left(1\right)\right] \\ \angle \mathrm{QRC} = \angle \mathrm{PRA} \left[\mathrm{vertically} \mathrm{opposite} \mathrm{angles}\right] \\ \Rightarrow △\mathrm{CQR} ~ △\mathrm{APR} \left[\mathrm{AA} \mathrm{similarity}\right] \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{\mathrm{CR}}{\mathrm{AR}} = \frac{\mathrm{CQ}}{\mathrm{AP}}=\frac{\mathrm{QR}}{\mathrm{PR}} \left(\mathrm{corresponding} \mathrm{sides} \mathrm{of} \mathrm{similar} △\text{'}\mathrm{s} \mathrm{are} \mathrm{proportional}\right) \\ \Rightarrow \frac{\mathrm{CR}}{\mathrm{AR}} = \frac{\mathrm{CQ}}{\mathrm{AP}} \\ \Rightarrow \frac{\mathrm{CR}}{\mathrm{AR}} =\frac{{\displaystyle \frac{4x}{5}}}{{\displaystyle \frac{3x}{5}}} \\ \Rightarrow \frac{\mathrm{CR}}{\mathrm{AR}} =\frac{4}{3} \\ \Rightarrow \frac{\mathrm{CR}}{\mathrm{AR}} + 1= \frac{4}{3} + 1 \\ \Rightarrow \frac{\mathrm{CR} + \mathrm{AR}}{\mathrm{AR}} = \frac{7}{3} \\ \Rightarrow \frac{\mathrm{AC}}{\mathrm{AR}} = \frac{7}{3} \\ \Rightarrow 7\mathrm{AR} = 3\mathrm{AC} \\ \Rightarrow \mathrm{AR} = \frac{3}{7}\mathrm{AC} \end{gathered}$$