Answer:
Area of rectangle APQB = Area of rectangle DPQC
Where:
PQ is the line passing through the intersection of the diagonals of the parallelogram ABCD
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.