Question

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

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Solution

## Given : $\text{PQ=12.5cm,PA=5cm,BR=6cm,PB=4cm}$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 : $\text{PQ=12.5cm,PA=5cm,BR=6cm,PB=4cm}$ (given)So $\text{QA=QP-PA=12.5-5=7.5cm}$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, $\text{AB∥QR}$.

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