Question

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

= half of (perimeter of ABC).​

Answer 1

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