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 and APQ is a straight line meeting BC at P and DC produced at Q.
To do:
We have to prove that the rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.
Solution:
In $\triangle ABP$ and $\triangle QDA$,
$\angle ABP=\angle QDA$ (Opposite angles of a parallelogram are equal)
$\angle BAP=\angle PQD$ (Alternate interior angles)
Therefore,
$\triangle ABP \sim\ \triangle QDA$ (By AA similarity)
This implies,
$\frac{AB}{QD}=\frac{BP}{DA}$ (Corresponding parts of similar triangles are proportional)
$\frac{AB}{QD}=\frac{BP}{BC}$ ($DA=BC$, opposite sides of a parallelogram are equal)
$AB \times BC = BP \times QD$
Therefore,
The rectangle obtained by BP and DQ is equal to the rectangle contained by AB and BC.
Hence proved.
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