Question

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

Answer 1

Step 1: Make the necessary constructions to the given triangle

It is given that $PS$ is the bisector of $\angle QPR$ of $△PQR$.

i.e., $\angle QPS=\angle SPR$ $...\left(\mathrm{i}\right)$

Now, draw a line $RX$ such that $RX\left|\right|SP$.

And, produce the line $QP$ such that it intersects $RX$ at point $T$.

Step 2: Determine the pair of equal angles

Since $RX\left|\right|SP$ and $RP$ is a transversal.

So, $\angle SPR=\angle PRT$ $...\left(\mathrm{ii}\right)$

And, since $RX\left|\right|SP$ and $QT$ is a transversal.

So, $\angle QPS=\angle PTR$ $...\left(\mathrm{iii}\right)$

Now, using the equation $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$, we have,

$\angle QPS=\angle PRT$ $...\left(\mathrm{iv}\right)$

Again, using the equation $\left(\mathrm{iii}\right)$ and $\left(\mathrm{iv}\right)$, we have,

$\angle PRT=\angle PTR$

Step 3: Use the property of isosceles triangles

As we know that the sides opposite to the equal angles of a triangle are also equal.

So, in $△PTR$,

$\because \angle PRT=\angle PTR$

$\Rightarrow$ $PT=PR$ $...\left(\mathrm{v}\right)$

Step 4: Use the basic proportionality theorem of triangles,

Now, using the basic proportionality theorem in $△QTR$, it can be stated that,

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

Using the equation $\left(\mathrm{v}\right)$ in the above, it can be stated that,

$\frac{QS}{PQ}=\frac{SR}{PR}$ $\left[\because QP=PQ\right]$

Which is the required answer.

Hence, it is proved that $\frac{QS}{PQ}=\frac{SR}{PR}$.