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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