A straight line is drawn through the intersection of the two diagonals of a parallelogram. Prove that it exactly divides the parallelogram into two equal parts by area.

Area of rectangle APQB = Area of rectangle DPQC

Where:

PQ is the line passing through the intersection of the diagonals of the parallelogram ABCD

Step-by-step explanation:

Here we note that a parallelogram is a quadrilateral with the two opposite sides equal, therefore, the diagonals each divides the parallelogram

Given the parallelogram ABCD with a point of intersection of the two diagonals = O

We are to prove that a line PQ passing through O divides the parallelogram into two equal parts;

The diagonals of a parallelogram bisect each other hence

OA = OC and OB = OD

Also ∠AOB = ∠COD (vertically opposite angles at the crossing of the diagonals)

∴ ΔAOB ≅ ΔCOD (SAS congruence rule)

Area of ΔAOB = Area of ΔCOD

In ΔAOP and ΔCOQ, we have;

∠PAO = ∠QCO (alternate interior angles of a parallel line)

OA = OC (as above)

∠AOP = ∠QOC (vertically opposite angles at the crossing of the diagonals)

∴ ΔAOP ≅ ΔQOC

Area of ΔAOP = Area of ΔQOC

From which we have by similarity;

ΔBOQ ≅ ΔPOD

Area of ΔBOQ = Area of  ΔPOD

Hence area of rectangle APQB = Area of ΔQOC + Area of ΔCOD + Area of  ΔPOD = Area of ΔAOP + Area of ΔAOB + Area of ΔBOQ

∴ Area of rectangle APQB = Area of rectangle DPQC there proved as required.

c - they will have the same distribution

step-by-step explanation:

this is the only one that could make sense. : )