(Q15) The diagonals of a quadrilateral ABCD intersect each other at point 'O' such that
AO
BO
 =
CO
DO
 . Prove that ABCD is a trapezium.

Solution :

ABCDO
Given:In quadrilateral ABCD ,
AO
BO
 =
CO
DO

RTP : ABCD is a trapezium

Construction : Through 'O' draw a line parallel to AB which meets DA at X

Proof :In ∆DAB, XO ∥ AB   (by construction)

⇒
DX
 =
OB
   (by B.P.T)
AX
XD
 =
....(1)
AO
 =
DO
   (given)
AO
 =
OD
 ....(2)

From (1) and (2),

AX
XD
 =
AO
CO
In ∆ADC, XO is a line such that
AX
XD
 =
AO
OC

⇒ XO ∥ DC   (by converse of the basic the proportionality theorem)

⇒ AB ∥ DC

In quadrilateral ABCD, AB ∥ DC

⇒ ABCD is a trapezium   (by definition)

Hence Proved.