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  • O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

Answers (1)

ABCD is a trapezium and O is the point of intersection of the diagonals AC and BD.

AB\parallel DC

Proof:- \text {In} \Delta ABD \; \text {and }\; \Delta POD

\angle D=\angle D               (Common angle)

\angle ABD=\angle POD        (corresponding angles)

\therefore \Delta ABD\sim \Delta POD     (by AA similarity criterion)

Then \frac{OP}{AB}=\frac{PD}{AD}\; \; \; \; \; \; \; \; \; \; ...(1)

\text {In} \Delta ABC \; \text {and }\; \Delta OQC

\angle C=\angle C               (Common angle)

\angle BAC=\angle QOC        (corresponding angles)

\therefore \Delta ABC\sim \Delta OQC     (by AA similarity criterion)

Then \frac{OQ}{AB}=\frac{QC}{BC}\; \; \; \; \; \; \; \; \; \; ...(2)

\text {In}\Delta ADC

OP\parallel DC

We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

\therefore \frac{AP}{PD}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(3)

\text {Also In}\; \Delta ABC

OQ\parallel AB

\therefore \frac{BQ}{QC}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(4)    (by basic proportionality theorem)

from equation (3) and (4)

\frac{AP}{PD}=\frac{BQ}{QC}

Add 1 on both sides we get

\frac{AP}{PD}+1=\frac{BQ+QC}{QC}

\frac{AP}{PD}=\frac{BC}{QC}

\Rightarrow \frac{PD}{AD}=\frac{QC}{BC}\; \; \; \; \; \; \; ...(5)

\frac{OP}{AB}=\frac{QC}{BC}                   (use equation (1))

\Rightarrow \frac{OP}{AB}=\frac{OQ}{AB}               (use equation (2))

\Rightarrow OP=OQ

Hence proved

Posted by

infoexpert23

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