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, `("OP")/("AB") = ("PD")/("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, `("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") = ("OA")/("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") = ("OQ")/("BC")`  ...[From equations (i) and (ii)]

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

⇒ OP = OQ

Hence proved.