Question

In a parallelogram ABCD, P divides AB in the ratio 2:5 and Q divides DC in the ratio 3:2. If AC and PQ intersect at R. Find the ratio AR:RC and PR:RQ.

Answer 1

ANSWER on Question #81273 – Math – Geometry

QUESTION

In a parallelogram $ABCD$, $P$ divides $AB$ in the ratio $2:5$ and $Q$ divides $DC$ in the ratio $3:2$. If $AC$ and $PQ$ intersect at $R$. Find the ratio $AR:RC$ and $PR:RQ$.

SOLUTION

Draw pattern to this problem

Since $P$ divides $AB$ in the ration $2:5$, then we introduce the proportionality coefficient $-x$.

$$AP = 2x \quad \text{and} \quad BP = 5x$$

Since $Q$ divides $DC$ in the ration $2:3$, then we introduce the proportionality coefficient $-y$.

$$CQ = 2y \quad \text{and} \quad DQ = 3y$$

Since $ABCD$ is a parallelogram, then

$$AB = CD \rightarrow AP + PB = CQ + DQ \rightarrow 2x + 5x = 2y + 3y \rightarrow 7x = 5y \rightarrow \boxed{\frac{x}{y} = \frac{5}{7}}$$

Consider triangles $\Delta APR$ and $\Delta CQR$:

$$\angle PRA = \angle CRQ \quad \text{as a pair of vertical angles}$$

(More information: https://en.wikipedia.org/wiki/Angle#Vertical_and_adjacent_angle_pairs)

$$\angle PAR = \angle QCR \quad \text{as a pair of internal multi-faceted angles with } AB \parallel CD \text{ and } AC - \text{secant}$$

Then,

$$\Delta A P R \sim \Delta C Q R \text{ triangles are similar (AAA, angle angle angle)}$$ $$\Delta A P R \sim \Delta C Q R \rightarrow \frac{A P}{C Q} = \frac{P R}{Q R} = \frac{A R}{C R} \rightarrow \frac{2 x}{2 y} = \frac{P R}{Q R} = \frac{A R}{C R} \rightarrow \frac{P R}{Q R} = \frac{A R}{C R} = \frac{x}{y} = \frac{5}{7}$$

Conclusion,

$$\frac{P R}{Q R} = \frac{5}{7} \rightarrow \boxed{P R : Q R = 5 : 7}$$ $$\frac{A R}{C R} = \frac{5}{7} \rightarrow \boxed{A R : C R = 5 : 7}$$

**ANSWER**

$$P R : Q R = 5 : 7$$ $$A R : C R = 5 : 7$$

Answer provided by https://www.AssignmentExpert.com