$ABCD$ is a parallelogram and $P$ and $Q$ are points on the diagonal $AC$ such that $AP = PQ = QC$. Prove that $BQ \parallel DP$ and $BD$ bisects $PQ$.
(N/A) $1$. In $\triangle ABP$ and $\triangle CDQ$:
$AB = CD$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle BAP = \angle DCQ$ (Alternate interior angles as $AB \parallel CD$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ABP \cong \triangle CDQ$.
Thus, $BP = DQ$ and $\angle ABP = \angle CDQ$.
$2$. In $\triangle ADP$ and $\triangle CBQ$:
$AD = CB$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle DAP = \angle BCQ$ (Alternate interior angles as $AD \parallel BC$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ADP \cong \triangle CBQ$.
Thus, $DP = BQ$ and $\angle ADP = \angle CBQ$.
$3$. Since $BP = DQ$ and $DP = BQ$, quadrilateral $PBQD$ is a parallelogram (opposite sides are equal).
Therefore, $BQ \parallel DP$.
$4$. In parallelogram $PBQD$, diagonals $BD$ and $PQ$ bisect each other. Hence, $BD$ bisects $PQ$.
$AB = CD$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle BAP = \angle DCQ$ (Alternate interior angles as $AB \parallel CD$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ABP \cong \triangle CDQ$.
Thus, $BP = DQ$ and $\angle ABP = \angle CDQ$.
$2$. In $\triangle ADP$ and $\triangle CBQ$:
$AD = CB$ (Opposite sides of parallelogram $ABCD$ are equal).
$\angle DAP = \angle BCQ$ (Alternate interior angles as $AD \parallel BC$ and $AC$ is a transversal).
$AP = CQ$ (Given).
By $SAS$ congruence criterion, $\triangle ADP \cong \triangle CBQ$.
Thus, $DP = BQ$ and $\angle ADP = \angle CBQ$.
$3$. Since $BP = DQ$ and $DP = BQ$, quadrilateral $PBQD$ is a parallelogram (opposite sides are equal).
Therefore, $BQ \parallel DP$.
$4$. In parallelogram $PBQD$, diagonals $BD$ and $PQ$ bisect each other. Hence, $BD$ bisects $PQ$.
Explore More
Similar Questions
Can all the angles of a quadrilateral be acute angles? Give reason for your answer.
Easy
View Solution$E$ is the mid-point of the side $AD$ of the trapezium $ABCD$ with $AB \parallel DC$. $A$ line through $E$ drawn parallel to $AB$ intersects $BC$ at $F$. Show that $F$ is the mid-point of $BC$.
Medium
View SolutionIn $\Delta ABC$,the points $M$ and $N$ are on sides $AB$ and $AC$ respectively such that $AM = \frac{1}{4} AB$ and $AN = \frac{1}{4} AC$. Prove that $MN = \frac{1}{4} BC$.
Medium
View Solution$P, Q, R$ and $S$ are respectively the mid-points of the sides $AB, BC, CD$ and $DA$ of a quadrilateral $ABCD$ such that $AC \perp BD$. Prove that $PQRS$ is a rectangle.
Difficult
View SolutionProve that the line segment joining the mid-points of the non-parallel sides of a trapezium is parallel to the parallel sides of the trapezium.
Difficult
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo