# Ex. 6.6 Q1 Triangles Solution - NCERT Maths Class 10

## Question

In below Fig, \(PS\) is the bisector of \(\angle {{QPR}}\) of \(\Delta PQR\). Prove that \(\begin{align}\frac{{QS}}{{SR}} = \frac{{PQ}}{{PR}}\end{align}\)

## Text Solution

**Reasoning:**

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.(BPT)

**Steps:**

Draw a line parallel to \(PS,\) through \(R,\) which intersect \(QP\) produced at \(T\)

\[PS\parallel RT\]

In \(\Delta QPR\)

\[\begin{align}&\angle QPS = \angle SPR \\ &\text{(PS is bisector of }\angle QPR \; \dots (1) \end{align}\]

But,

\[\begin{align}& \angle PRT = \angle SPR \\ &\text{(Alternate interior angles)}\; \dots (2) \end{align}\]

\[\begin{align}& \angle QPS = \angle PTR \\ &\text{(Corresponding angles)}\; \dots (3) \end{align}\]

From \(\rm{(1), (2)}\) and \(\rm{(3)}\)

\[\begin{align}\angle PTR &= \angle PRT\\ PR &= PT\;.......\rm (4)\end{align}\]

(Since in a triangle, sides opposite to the equal angles are equal)

In \(\Delta QSP,PS \parallel RT\)

\[\begin{align}\frac{{QS}}{{SR}}&= \frac{{QP}}{{PT}}\;\;\;\left[ {{\rm{BPT}}} \right]\\\frac{{QS}}{{SR}}&= \frac{{QP}}{{PR}}\;\;\;\left[ {{\rm{from (4)}}} \right]\end{align}\]