In $\square ABCD$,$P \in \overline{AB}$,$Q \in \overline{BC}$,$R \in \overline{CD}$,and $S \in \overline{AD}$ such that $\frac{AP}{AB} = \frac{CQ}{BC} = \frac{CR}{CD} = \frac{AS}{AD} = \frac{1}{3}$. Prove that $\square PQRS$ is a parallelogram.
(A) Given: $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$.
In $\triangle ABD$,by the Converse of Basic Proportionality Theorem $(BPT)$,since $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$,it implies $\frac{AP}{PB} = \frac{AS}{SD} = \frac{1}{2}$. Thus,$PS \parallel BD$ and $PS = \frac{1}{3} BD$.
Similarly,in $\triangle BCD$,since $\frac{CQ}{BC} = \frac{CR}{CD} = \frac{1}{3}$,we have $\frac{CQ}{QB} = \frac{CR}{RD} = \frac{1}{2}$. Thus,$QR \parallel BD$ and $QR = \frac{1}{3} BD$.
From these two results,$PS \parallel QR$ and $PS = QR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$\square PQRS$ is a parallelogram.
In $\triangle ABD$,by the Converse of Basic Proportionality Theorem $(BPT)$,since $\frac{AP}{AB} = \frac{AS}{AD} = \frac{1}{3}$,it implies $\frac{AP}{PB} = \frac{AS}{SD} = \frac{1}{2}$. Thus,$PS \parallel BD$ and $PS = \frac{1}{3} BD$.
Similarly,in $\triangle BCD$,since $\frac{CQ}{BC} = \frac{CR}{CD} = \frac{1}{3}$,we have $\frac{CQ}{QB} = \frac{CR}{RD} = \frac{1}{2}$. Thus,$QR \parallel BD$ and $QR = \frac{1}{3} BD$.
From these two results,$PS \parallel QR$ and $PS = QR$.
Since one pair of opposite sides of quadrilateral $PQRS$ is equal and parallel,$\square PQRS$ is a parallelogram.
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Similar Questions
Which of the following correctly matches the information in Part $I$ and Part $II$?
Part $I$ Part $II$ $1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median. $a. AB^2 + BC^2 = 2(BD^2 + CD^2)$ $2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude. $b. BC = \frac{1}{2} AB$ $3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$. $c. AC^2 = CD \cdot BC$ $4.$ In $\Delta ABC$,$\overline{BD}$ is a median. $d. BM = \frac{1}{2} AC$
| Part $I$ | Part $II$ |
|---|---|
| $1.$ In $\Delta ABC$,$\angle B$ is a right angle and $\overline{BM}$ is a median. | $a. AB^2 + BC^2 = 2(BD^2 + CD^2)$ |
| $2.$ In $\Delta ABC$,$\angle A$ is a right angle and $\overline{AD}$ is an altitude. | $b. BC = \frac{1}{2} AB$ |
| $3.$ In $\Delta ABC$,$m\angle C = 90^\circ$ and $m\angle A = 30^\circ$. | $c. AC^2 = CD \cdot BC$ |
| $4.$ In $\Delta ABC$,$\overline{BD}$ is a median. | $d. BM = \frac{1}{2} AC$ |
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