Question

In figure, $PS$ is the bisector of $\angle QPR$ of $△PQR$. Prove that
$\frac{QS}{SR}=\frac{PQ}{PR}$

Answer 1

In $△QRT$

$RT||SP$ [By construction]

And PS intersects QT and QR at two distinct points P and Q

Therefore,

QT and QR will be divided in the same ratio

$\frac{QS}{SR}=\frac{PQ}{PT}......\left(1\right)$

[Basic proportionality theorem : If a line is drawn parallel to one side of a triangle, intersecting other two sides at distinct points, then other two sides are divided in the same ratio.]

Now,

$RT||SP$

and PR is the transversal

Therefore,

$\angle SPR=\angle PRT....\left(2\right)$ [Alternate interior angles]

and

$\angle QPS=\angle PTR....\left(3\right)$ [Corresponding angles]

Also, given the

PS is the bisector of $\angle QPR$

$\angle QPR=\angle SPR$

From (2) and (3)

$\angle PTR=\angle PRT$

Therefore, $PT=PR$ [Sides opposite to equal angles of a trinagle are equal]

Putting PT=PR in (1)

$\frac{QS}{SR}=\frac{PQ}{PT}$

Hence proved.