Question

Question 15
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 in P and BC in Q. Prove that PO = QO.

Answer 1

Given, ABCD is a trapezium, the diagonals AC and BD intersect at O.
Also, PQ ||AB||DC.
To prove : PO = QO.

Proof :

$In\Delta ABDand\Delta POD$

$PO||AB[\because PQ||AB]$

$\angle D=\angle D\left[commonangle\right]$
$\angle ABD=\angle POD\left[correspondingangles\right]$
$\therefore \Delta ABD\sim \Delta POD\left[byAAAsimilaritycriterion\right]$

$Then,\frac{OP}{AB}=\frac{PD}{AD}\dots \left(i\right)$

$In\Delta ABCand\Delta OQC$,

$OQ||AB[\because PQ||AB]$

$\angle C=\angle C\left[commonangle\right]$
$\angle BAC=\angle QOC\left[correspondingangles\right]$
$\therefore \Delta ABC\sim \Delta OQC\left[byAAAsimilaritycriterion\right]$
$Then,\frac{OQ}{AB}=\frac{QC}{BC}\dots \left(ii\right)$

$Now,in\Delta ADC$,
$OP||DC[\because PQDC]$
$\therefore \frac{AP}{PD}=\frac{OA}{OC}\left[bybasicproportionallytheorem\right]...\left(iii\right)$

$In\Delta ABC$,
$OQ||AB[\because PQ||AB]$
$\therefore \frac{BQ}{QC}=\frac{OA}{OC}$
[by basic proportionally theorem] ...(iv)

From Eq.(iii) and (iv),
Adding 1 on both sides, we get,
$\frac{AP}{PD}+1=\frac{BQ}{QC}+1$
$\Rightarrow \frac{AP+PD}{PD}=\frac{BQ+QC}{QC}$
$\Rightarrow \frac{AD}{PD}=\frac{BC}{QC}$
$\Rightarrow \frac{PD}{AD}=\frac{QC}{BC}$
$\Rightarrow \frac{OP}{AB}=\frac{OQ}{AB}[fromEqs.(i\left)and\right(ii\left)\right]$
$\Rightarrow OP=OQ$