In the given figure, the incircle of ABCtouches the sides BC, CA and A

1.	In the given figure, the incircle of ABCtouches the sides BC, CA and AB at P, Q andRrespectively. Prove that(AR + BP+CQ) = (AQ + BR+CP)= half of (perimeter of ABC).
Answer» Answer: As SEEN in the figure:- AR=AQ (tangents from a point to a circle are always equal in LENGTH)......................(i) BP =BR (tangent from point B to circle...so...BP same in LENGTH as BR)..............(ii) CQ=CP ( similar REASON)..............(iii) ----------- adding the 3 equations:- (lhs=rhs) AR+BP+CQ=AQ+BR+CP From the figure we can see that:- perimeter=AC+CB+BA Further breaking this equation:- =(AQ+QC)+(CP+PB)+(BR+RA) =》2(AQ+BR+CP)=2(AR+BP+CQ)= Perimeter [from(i),(ii), (iii)] therefore, (AQ+BR+CP)=(AR+BP+CQ)= ½ (PERIMETER) *HENCE* *PROVED*

In the given figure, the incircle of ABCtouches the sides BC, CA and AB at P, Q andRrespectively. Prove that(AR + BP+CQ) = (AQ + BR+CP)= half of (perimeter of ABC).

Answer»

Answer:

As SEEN in the figure:-

AR=AQ (tangents from a point to a circle are always equal

in LENGTH)......................(i)

BP =BR (tangent from point B to circle...so...BP same in

LENGTH as BR)..............(ii)

CQ=CP ( similar REASON)..............(iii)

-----------

adding the 3 equations:-

(lhs=rhs)

AR+BP+CQ=AQ+BR+CP

From the figure we can see that:-

perimeter=AC+CB+BA

Further breaking this equation:-

=(AQ+QC)+(CP+PB)+(BR+RA)

=》2(AQ+BR+CP)=2(AR+BP+CQ)= Perimeter [from(i),(ii),

(iii)]

therefore, (AQ+BR+CP)=(AR+BP+CQ)= ½ (PERIMETER)

HENCE PROVED