Question

In fig., \[PS\] is the bisector \[\angle QPR\] of \[\Delta PQR\]. Prove that, \[\dfrac{{QS}}{{SR}} = \dfrac{{PQ}}{{PR}}\].
               

Answer 1

Hint: We need to draw a construction to proof the given problem.
Using the properties of parallel lines, we can find the relation between the angles.
By using proportionality theorem we can prove the given problem.

Complete step-by-step answer:
It is given that; \[PS\] is the bisector \[\angle QPR\] of \[\Delta PQR\].
We have to prove that \[\dfrac{{QS}}{{SR}} = \dfrac{{PQ}}{{PR}}\].
               

We will draw a line from the point \[Q\] which is parallel to \[PS\] and it meets the side \[PR\] at \[T\]. So, we have, \[PS\parallel TQ\].
We know that the alternate angles of the parallel line are equal.
Since, \[PS\parallel TQ\] and \[PQ\] is the transversal. Then, \[\angle TQP = \angle QPS\]… (1), since they are alternate angles.
Also, \[\angle QPS = \angle PTQ\]… (2) as they are corresponding angles.
Again, we know that, \[PS\] is the bisector of \[\angle QPR\].
So, \[\angle QPS = \angle RPS\]
Therefore, from (1) and (2) we get,
\[\angle TQP = \angle PTQ\]
Here, \[\Delta PTQ\] is a triangle whose \[\angle TQP = \angle PTQ\].
So, the opposite sides are equal.
That means, \[PT = PQ\]… (3)
Proportionality theorem states that, if a line is drawn parallel to any one side of a triangle in such a way that it intersects the other two sides in two distinct points then the other two sides of the triangle are divided in the same ratio.
In the \[\Delta RTQ\], \[PS\parallel TQ\], by the proportionality theorem, we get,
\[\dfrac{{QS}}{{SR}} = \dfrac{{TP}}{{PR}}\]
From (3) we get,
\[\dfrac{{QS}}{{SR}} = \dfrac{{PQ}}{{PR}}\]
Here’s the proof.

Note: In an isosceles triangle, the opposite sides of the equal angles are equal.
Proportionality theorem states that, if a line is drawn parallel to any one side of a triangle in such a way that it intersects the other two sides in two distinct points then the other two sides of the triangle are divided in the same ratio.