Ex. 6.6 Q1 Triangles Solution - NCERT Maths Class 10

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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


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)


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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