Q. 15

# 0 is the point of

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation view all courses 