Question

Prove that, in a triangle, the angle bisector divides the side opposite to the angle in the ratio of the remaining sides.

Answer 1

Given: In $△PQR$, ray $RT$ bisects $PRQ$

To prove: $\frac{PT}{TQ}=\frac{PR}{RQ}$

Construction: Draw a ray from $Q$ parallel to ray $RT$ such that it intersects extended $PR$ at $S$.

Proof:

We have,

$PR||QS$ and $PS$ is transversal

$\therefore$ $PRT=RSQ$ ....(1) (Corresponding angles)

Corresponding other transversal $RQ$, we obtain

$TRQ=RQS$ ....(2) Alternate angles

But $PRT=TRQ$ ....$(RT$ bisects $PRQ)$

$\therefore ,RSQ=RQS$ ....From (1)and (2)

Thus, in $△RQS$,

$RS=RQ$ ....(3) (Sides opposite to equal angles are equal)

Now, in $△PQS$, we have

$\frac{PT}{TQ}=\frac{PR}{RS}$ ....ByBPT

$\Rightarrow \frac{PT}{TQ}=\frac{PR}{RQ}$ ....Using (3)

Hence, proved.