Question

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.

Answer 1

Given ABCD is a trapezium.

Diagonals AC and BD are intersect at O.

PQ || AB || DC

To prove: PO = QO

Proof: In ∆ABD and ∆POD,

PO || AB  ...[∵ PQ || AB]

∠D = ∠D   ...[Common angle]

∠ABD = ∠POD  ...[Corresponding angles]

∴ ∆ABD ~ ∆POD  ...[By AAA similarity criterion]

Then, ${\displaystyle \frac{\text{OP}}{\text{AB}}=\frac{\text{PD}}{\text{AD}}}$  ...(i)

In ∆ABC and ∆OQC,

OQ || AB  ...[∵ OQ || AB]

∠C = ∠C  ...[Common angle]

∠BAC = ∠QOC  ...[Corresponding angles]

∴ ∆ABC ~ ∆OQC  ...[By AAA similarity criterion]

Then, ${\displaystyle \frac{\text{OQ}}{\text{AB}}=\frac{\text{QC}}{\text{BC}}}$  ...(ii)

Now, In ∆ADC,

OP || DC

∴ ${\displaystyle \frac{\text{AP}}{\text{PD}}=\frac{\text{OA}}{\text{OC}}}$  [By basic proportionality theorem] ...(iii) 

In ∆ABC,

OQ || AB

∴ ${\displaystyle \frac{\text{BQ}}{\text{QC}}=\frac{\text{OA}}{\text{OC}}}$ [By basic proportionality theorem] ...(iv) 

From equations (iii) and (iv), we get

${\displaystyle \frac{\text{AP}}{\text{PD}}=\frac{\text{BQ}}{\text{QC}}}$

Adding 1 on both sides, we get

${\displaystyle \frac{\text{AP}}{\text{PD}}+1=\frac{\text{BQ}}{\text{QC}}+1}$

⇒ ${\displaystyle \frac{\text{AP}+\text{PD}}{\text{PD}}=\frac{\text{BQ}+\text{QC}}{\text{QC}}}$

⇒ ${\displaystyle \frac{\text{AD}}{\text{PD}}=\frac{\text{BC}}{\text{QC}}}$

⇒ ${\displaystyle \frac{\text{PD}}{\text{AD}}=\frac{\text{QC}}{\text{BC}}}$

⇒ ${\displaystyle \frac{\text{OP}}{\text{AB}}=\frac{\text{OQ}}{\text{BC}}}$  ...[From equations (i) and (ii)]

⇒ ${\displaystyle \frac{\text{OP}}{\text{AB}}=\frac{\text{OQ}}{\text{AB}}}$  ...[From equation (ii)]

⇒ OP = OQ

Hence proved.