Question

ABCD is a trapezium with AB║CD. P and Q are the midpoints of diagonals AC and BD. Prove that PQ║AB║CD and PQ = 1/2(AB-DC).

Answer 1

Answer:

In △ABD, PO∥AB [∵ PQ∥AB ]

⇒

AP

DP

=

OB

DO

[ By basic proportionality theorem ] ------ ( 1 )

⇒ In △BDC, OQ∥DC [

∵ PQ∥DC ]

QC

BQ

=

OD

OB

[ By basic proportionality theorem ]

⇒

BQ

QC

=

OB

OD

----- ( 2 )

⇒

AP

DP

=

BQ

QC

[ From ( 1 ) and ( 2 ) ]

⇒

AP

18

=

35

15

∴ AP=

15

18×35

=42

∴ AD=AP+DP=42+18=60cm