ABCD is a parallelogram and APQ is a straight-line meeting BC at P and DC produced at Q. Prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.

Given: ABCD is a parallelogram

To prove: BP x DQ = AB x BC

Theorem Used:

1) If two corresponding angles of two triangles are equal the triangles are said to be similar.

2) Basic proportionality theorem:

If a line is drawn parallel to one side of a triangle intersecting the other two sides, then it divides the two sides into same ratio.

Proof:

In ΔABP and ΔQDA

∠B = <D (Opposite angles of parallelogram)

∠BAP = ∠AQD (Alternative interior angle)

Then, ΔABP ~ ΔQDA

SO, (Corresponding parts of similar triangle area proportion)

But, DA = BC (Opposite side of parallelogram)

Then,

Or, AB x BC = QD X BP