the incircle of a triangle abc touches the sides bc, ca and ab at p ,q and r respectively. prove that (ar+bp+cq)=(aq+br+cp)=1/2 of perimeter of abc

Given that the incircleof a triangle touches the sides BC, CA, and AB at P, Q and R respectively.
To prove that (AR+ BP+CQ)= (AQ+BR+CP)= 1/2* perimeter of triangle ABC.
Proof: In triangle ABC,
AR=AQ - (tangents from an exterior point are equal.) -(1)
BP=BR -(tangents from an exterior point are equal.)​ -(2)
CQ=CP -(tangents from an exterior point are equal.) -(3)
By adding (1),(2) and (3).
(AR+BP+CQ)=(AQ+BR+CP)
Now,
AR= 1/2 AB
BP= 1/2 BC
CQ= 1/2 AC
By adding them,
(AR+BP+CQ)= AB+BC+AC/2= 1/2 of perimeter of triangle ABC
Similarly, (AQ+BR+CP)= 1/2 of perimeter of triangle ABC
Therefore,​(AR+ BP+CQ)= (AQ+BR+CP)= 1/2* perimeter of triangle ABC.
Hence, proved.