Question

if the diagonals of a quadrilateral abcd intersect each other at 90 degree a,then prove that the quadrilateral form by joong the midpoints of the adjustments sides is a rectangle​

Answer 1

Answer:

Given:

The diagonals of a quadrilateral ABCD intersect each other at the point O such that

AO/BO = CO/DO

i.e.,

AO/CO = BO/DO

To Prove: ABCD is a trapezium

Construction:

Draw OE∥DC such that E lies on BC.

Proof:

In △BDC,

By Basic Proportionality Theorem,

BO/OD = BE/EC ...(1)

But,

AO/CO = BO/DO (Given) ...(2)

∴ From (1) and (2)

AO/CO = BE/EC

Hence, By Converse of Basic Proportionality Theorem,

OE∥AB

Now Since, AB ∥ OE ∥ DC

∴ AB ∥ DC

Hence, ABCD is a trapezium.