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

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## 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 $$\Delta QPR$$
$$\angle QPS = \angle SPR$$  (Since PS is the bisector of $$\angle QPR$$) ...(i)
But $$\angle PRT = \angle SPR$$ (Alternate interior angles)....(ii)
$$\angle QPS = \angle PTR$$ (Corresponding angles)...(ii)

From (i), (ii) and (iii) \begin{align}\angle PTR &= \angle PRT\\ PR &= PT\;.......(iv)\end{align}
(Since in a triangle, sides opposite to the equal angles are equal)

In $$\Delta QSP,PS \parallel RT$$

$\begin{array}{l}\frac{{QS}}{{SR}} = \frac{{QP}}{{PT}}\;{\rm{ }}\left[ {{\rm{BPT}}} \right]\\\frac{{QS}}{{SR}} = \frac{{QP}}{{PR}}\;{\rm{ }}\left[ {{\rm{from (iv)}}} \right]\end{array}$

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