Answer :

**Given:**

ABCD is a trapezium,

Diagonals AC and BD are intersect at 0.

**To prove:** PQ || AB || DC.

PO = QO**Concepts Used:**

AAA Similarity Criterion: If all three angles of a triangle equals to angles of another triangle, then both the triangles are similar.

Basic Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

**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) [by basic proportionality theorem]

In ∆ABC and ∆OQC,

OQ || AB [∵ OQ || AB]

∠C = ∠C [common angle]

∠BAC = ∠QOC [corresponding angle]

∴ ∆ABC ∼ ∆OQC [by AAA similarity criterion]

Then,

OQ/AB = QC/BC …(ii) [by basic proportionality theorem]

Now, in ∆ADC,

OP || DC

∴ AP/PD = 0A/0C [by basic proportionality theorem] …(iii)

In ∆ABC, OQ || AB

∴ BQ/QC = OA/OC [by basic proportionality theorem] …(iv)

From Equation (iii) and (iv),

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 Equation (i) and (ii)]

⇒ OP/AB = OQ/AB [from Equation (iii)]

⇒ OP = OQ

**Hence proved.**

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