Question

ABCD is a parallelogram, prove that 

1. Area (ΔABC) = Area (ΔPAD)
2. Area (ΔPAB) = Area (quadrilateral ACPD)

Answer 1

Given:

ABCD is a parallelogram.

P is any point (assumed inside or on the parallelogram unless otherwise stated).

i. Prove that Area(ΔABC) = Area(ΔPAD)

Let’s analyze both triangles:

• ΔABC lies on base BC and between vertices A and C.
• ΔPAD lies on base AD and between vertices P and D.

But to prove Area(ΔABC) = Area(ΔPAD), a common trick is to observe their relative positions in the parallelogram.

Let’s assume point P lies on diagonal BD, which divides the parallelogram into two equal areas.

• Then, Area(ΔABC) and Area(ΔPAD) are on opposite halves of the parallelogram and share the same base length and height (as they lie between the same set of parallel lines).

• Since diagonal divides parallelogram into two equal areas and P lies on diagonal BD,

Thus, Area(ΔABC) = Area(ΔPAD).

ii. Prove that Area(ΔPAB) = Area(quadrilateral ACPD)

• The triangle PAB and quadrilateral ACPD together make up the entire parallelogram ABCD.

We already established in (i) that:

• Area(ΔABC) = Area(ΔPAD)

So, parallelogram ABCD can be split as:

Area(ΔPAB) + Area(ΔPAD) = Area(ABCD)Area(ΔABC) + Area(quadrilateral ACPD) = Area(ABCD)

Since Area(ΔPAD) = Area(ΔABC), it implies Area(ΔPAB) = Area(quadrilateral ACPD).