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.

#### Solution

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 AA similarity criterion]

Then, `(OP)/(AB) = (PD)/(AD)` .......(i)

In ∆ABC and ∆OQC, OQ || AB

∠C = ∠C ......[Common angle]

∠B AC = ∠QOC ......[Corresponding angles]

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

Then, `(OQ)/(AB) = (QC)/(BC)` ......(ii)

Now, In ∆ADC, OP || DC

∴ `(AP)/(PD) = (OA)/(OC)` [By basic proportionality theorem] .......(iii)

In ∆ABC, OQ || AB

∴ `(BQ)/(QC) = (OQ)/(OC)` [By basic proportionality theorem] ......(iv)

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

`(AP)/(PD) = (BQ)/(QC)`

Adding 1 on both sides, we get

`(AP)/(PD) + 1 = (BQ)/(QC) + 1`

⇒ `(AP + PD)/(PD) = (BQ + QC)/(QC)`

⇒ `(AD)/(PD) = (BC)/(QC)`

⇒ `(PD)/(AD) = (QC)/(BC)`

⇒ `(OP)/(AB) = (QC)/(BC)` ....[From equation (i)]

⇒ `(OP)/(AB) = (OQ)/(AB)` .....[From equation (ii)]

⇒ OP = OQ

Hence proved