Question

O is the point of intersection of diagonals of trapezium ABCD. 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

Answer:

OP = OQ.

Step-by-step explanation:

See the attached diagram for the description of the question.  

ABCD is a trapezoid with AB || CD

AC and BD are diagonals which will meet at point O.

A line PQ is drawn from O which is parallel to AB and CD.  PQ || AB || CD

In ∆ABD and ∆POD

AB || PO

∠ADB = ∠ PDO (Since they are same angles)

∠ABD = ∠POD (Since they are corresponding angles on the parallel lines)

∠DAB = ∠DPO (Corresponding angles)

By AAA law, we can say that  ∆ABD ~ ∆POD

From above similar triangles, we can get  

OP/AB = PD/AD  -----------------------------------------------------------E1

In ∆ABC and ∆OQC

AB || OQ

∠ACB = ∠OCQ

∠BAC = ∠QOC

∠ABC = ∠OQC.

BY AAA law, we can say that ∆ABC ~ ∆OQC

We can derive  

OQ/AB = CQ/BC  ----------------------------------------------------------E2

BQ/CQ = OA/OC-----------------------------------------------------------E3

Similarly we can prove that ∆ADC ~ ∆APO

We can derive  

AP/PD = OA/OC ---------------------------------------------------------E4

From E3 and E4, we can derive

AP/PD = BQ/CQ

Add + 1 both sides, we get

AP/PD + 1 = BQ/CQ + 1

(AP+PD)/PD = (BQ+CQ)/CQ

AD/PD = BC/CQ

PD/AD = CQ/BC

From E1 and E2 and above result, we can derive

OP/AB = OQ/AB  

Hence OP = OQ.