In the following figure; IA and IB are bisectors of angles CAB and CBA respectively. CP is parallel to IA and CQ is parallel to IB. Then we can say that PQ=the perimeter of the $\Delta ABC$.

A.True
B.False

Hint: Use the theorem that the corresponding angles for parallel lines are always equal to show that angles corresponding to line CP and IA are equal. Then use the exterior angle property which states that the sum of any two interior angles is equal to the exterior angle. Then we know that the sides opposite to equal angles are equal so use it to prove that AC=AP and BC=BQ. Then we can write PQ as the sum of AP, AB and BQ.

Complete step-by-step answer:
Given, In$\Delta ABC$, IA and IB are bisectors of angles CAB and CBA respectively. Also, CP is parallel to IA and CQ is parallel to IB.
We have to prove that PQ= the perimeter of the$\Delta ABC$
Now since IA is angle bisector of angle CAB then,
$ \Rightarrow \angle {\text{CAB = }}\angle {\text{IAB}}$ --- (i)
Also CP is parallel to IA then we have,
$\angle {\text{IAB = }}\angle {\text{CPA}}$--- (ii) {As they are corresponding angles for parallel lines}
Now we know from exterior angle property that the sum of any two interior angles is equal to the exterior angle
Then we can write,
$ \Rightarrow \angle {\text{CAB = }}\angle {\text{CPA + }}\angle {\text{ACP}}$
Now we know that we can write angle CAB= sum of angles CAI and angle IAB
Then we have,
$ \Rightarrow \angle {\text{CAI + }}\angle {\text{IAB = }}\angle {\text{CPA + }}\angle {\text{ACP}}$
On substituting value from eq. (ii) we get,
$ \Rightarrow \angle {\text{CAI + }}\angle {\text{IAB = }}\angle {\text{IAB + }}\angle {\text{ACP}}$
On cancelling similar terms we get,
$ \Rightarrow \angle {\text{CAI = }}\angle {\text{ACP}}$ --- (iii)
Then from eq. (i) (ii) and (iii) we get,
$ \Rightarrow \angle {\text{CAI = }}\angle {\text{IAB = }}\angle {\text{CPA = }}\angle {\text{ACP}}$ --- (iv)
Now since IB is angle bisector of angle CBA then following the same process we can write,
$ \Rightarrow \angle {\text{CBI = }}\angle {\text{IBA = }}\angle {\text{BCQ = }}\angle {\text{BQC}}$ --- (v)
Now in $\Delta {\text{ACP}}$
$ \Rightarrow \angle {\text{CPA = }}\angle {\text{ACP}}$ {From eq. (iv)}
Then AC=AP --- (vi) {as sides opposite to equal angles are also equal}
 Then in $\Delta {\text{BCQ}}$ also,
$\angle {\text{BCQ = }}\angle {\text{BQC}}${From eq. (v)}
Then BC=BQ--- (vii)
Now we can write -
$ \Rightarrow $ PQ=AP+AB+BQ
Then substituting values from eq. (vi) and (vii), we have,
$ \Rightarrow $ PQ=AC+AB+BC
Since we know that perimeter is the sum of all three sides of the triangle then we can write ,
$ \Rightarrow $ PQ=perimeter of $\Delta ABC$
Hence the correct answer is A.

Note: Here you can also solve the given question by first finding the perimeter of triangle ABC which is given as-
Perimeter of $\Delta ABC$=AB+BC+CA--- (i)
And then using angle bisector theorem and exterior angle property prove that AC=AP and BC=BQ as proven in the above question. Then on substituting these values in eq. (i) you will see the Perimeter of $\Delta ABC$=AP+AB+BQ. And from diagram it is clear that-
 PQ=AP+AB+BQ
Hence, PQ=perimeter of $\Delta ABC$.