O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB`abs()`DC. Through O , aline segment PQ is drawn parallel to AB meeting AD in P and BC in Q, prove that PO=QO.

Explanation

Given ABCD is a trapezium, Diagonals AC and BD are intersect at O .
`PQabs()ABabs()DC`
To prove PO=QO

Proof in `DeltaABD and DeltaPOD, POabs()AB` [`becausePQabs()AB`]
`angleD=angleD` [common angles]
`angleABD=anglePOD` [corresponding angles]
`therefore DeltaABD~DeltaPOD` [by AAA similarity criterion]
Then, `(OP)/(AB)=(PD)/(AD)`...(i)
In `DeltaABC and DeltaOQC, OQabs()AB` [`because OQabs()AB`]
`angleC=angleC` [common angle]
`angleBAC=angleQOC` [corresponding angles]
`thereforeDeltaABC~DeltaOQC` [by AAA similarity criterion]
Then, `(OQ)/(AB)=(QC)/(BC)`..(ii)
Now in `DeltaADC`OP`abs()`DC
`therefore (AP)/(PP)=(OA)/(OC)` [by basic propertionality theorem]...(iii)
In `DeltaABC" "OQabs()AB`
`therefore (BQ)/(QC)=(OA)/(OC)` [by basic propertionality theorem]....(iv)
From Eqs. (iii) and (iv),
`(AP)/(PD)=(BQ)/(QC)`
Adding 1 on both sides, we get
`(AP)/(PD)+1=(BQ)/(QC)+1`
`rArr (AP+PD)/(PD)=(BQ+QC)/(QC)`
`rArr (AD)/(PD)=(BC)/(QC)`
`rArr (PD)/(AD)=(QC)/(BC)`
`rArr (OP)/(AB)=(OQ)/(BC)` [from Eqs. (i) and (ii)]
`rArr (OP)/(AB)=(OQ)/(AB)` [from Eq. (i)]
`rArr OP=OQ` Hence proved