Question

In the figure, $PR\parallel BC$ and $QR\parallel BD$. Prove that $PQ||CD$.

Answer 1

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

The converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

In the given figure, $PR||BC$ and $QR||BD$

In $△ABC$, $PR||BC$

Using the basic proportionality theorem, we have

$\frac{AP}{PC}=\frac{AR}{RB}\dots \left(1\right)$

Now, in $△ABD$, $QR||BD$

Again using the basic proportionality theorem, we have

$\frac{AQ}{QD}=\frac{AR}{RB}\dots \left(2\right)$

Comparing equations $\left(1\right)$ and $\left(2\right)$ we get,

$\frac{AP}{PC}=\frac{AQ}{QD}$

Now on applying this relation to $△ADC$, then using the converse of basic proportionality theorem we get $PQ||DC$.

Hence, $PQ||CD$.