Question

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

Answer 1

Given, ABCD is a trapezium, Diagonals AC and BD are intersect at O.
PQ II AB II DC.
To prove PO = QO


Proof In $\Delta ABD$ and $\Delta POD$ $POIIAB$ $[\because PQIIAB]$
$\angle D=\angle D$ [common angle]
$\angle ABD=\angle POD$ [corresponding angles]
$\therefore$ $\Delta ABD\sim \Delta POD$ [by AAA similarity criterion]
Then, $\frac{OP}{AB}=\frac{PD}{AD}$ ……(i)
In $\Delta ABC$ and $\Delta OQC$, $OQIIAB$ $[\because OQIIAB]$
$\angle C=\angle C$ [common angle]
$\angle BAC=\angle QOC$ [corresponding angles]
$\therefore$ $\Delta ABC\sim \Delta OQC$ [by AAA similarity criterion]
Then, $\frac{OQ}{AB}=\frac{QC}{BC}$ ……(ii)
Now, in $\Delta ADC$, $OPIIDC$
$\therefore$ $\frac{AP}{PP}=\frac{OA}{OC}$ [by basic proportionally theorem] …..(iii)
In $\Delta ABC$ $OQIIAB$
$\therefore$ $\frac{BQ}{QC}=\frac{OA}{OC}$ [by basic proportionally theorem] …..(iv)
From Eq.(iii and (iv),
Adding 1 on both sides, we get
$\frac{AP}{PD}+1=\frac{BQ}{QC}+1$
$\Rightarrow \frac{AP+PD}{PD}=\frac{BQ+QC}{QC}$
$\Rightarrow \frac{AP}{PD}=\frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD}=\frac{QC}{BC}$
$\Rightarrow \frac{OP}{AB}=\frac{OQ}{BC}$ [from Eqs.(i) and (ii)]
$\Rightarrow \frac{OP}{AB}=\frac{OQ}{AB}$ [from Eq.(ii)]
$\Rightarrow OP=OQ$