Question

$\text{A}$ and $\text{B}$ are respectively the points on the sides $\text{PQ}$ and $\text{PR}$ of a $△\text{PQR}$ such that $\text{PQ=12.5 cm, PA=5 cm, BR=6 cm}$ and $\text{PB=4 cm}$. Is $\text{AB∥QR}$ ? Give reasons for your answers ?

Answer 1

Given : $\text{PQ=12.5 cm , PA=5 cm , BR=6 cm , PB=4 cm}$

To prove : $\text{AB∥QR}$

We will solve it using Converse of Basic Proportionality Theorem .

Converse of Basic Proportionality Theorem : If a line segment intersecting two sides in two distinct points divides them in the same ratio then a line is parallel to the third side .

Proof : $\text{PQ=12.5 cm ,PA=5 cm ,BR=6 cm ,PB=4 cm}$ (given)

So $\text{QA=QP-PA=12.5-5=7.5 cm}$

Now we find the ratio of

$\frac{\text{PA}}{\text{AQ}}$$=\frac{5}{7.5}=\frac{50}{75}=\frac{2}{3}$

Now we find the ratio of

$\frac{\text{PB}}{\text{BR}}=\frac{4}{6}=\frac{2}{3}$

Here $\frac{\text{PA}}{\text{AQ}}=\frac{\text{PB}}{\text{BR}}=\frac{2}{3}$

So by using the converse of BPT we can say that $\text{AB∥QR}$ .

Hence, $\text{AB∥QR}$.