Triangle ABC hsa its base on line segment PN and vertex A on line PM. Circles with centres O and Q, having radii r1 and r2, respectively, are tangent to the triangle ABC externally at point K and L and to each of PM and PN.
(a) Prove that the line through K and L cuts the perimeter of triangle ABC into two equal pieces.
(b)Let T be the point of contact of BC with the circle inscribed in triangle ABC. Prove that (TC)(r1)+(TB)(r2) is equal to the area of triangle ABC.

Proof:
(a) KB=EB, LC=CD, AK=AF, AL=AG, so KB+BT+TC+CL = ED = FG =AK+AL. This means K and L cut triangle perimeter in half.

(b) Area OEDQGF = ED*(r1+r2), Area ABC = perimeter*r/2 = ED*r ,
So, OEDQGF=ABC*(r1+r2)/r
Area ACDQG = 2*ACQ=AC*r2,
area ABEOF=2*ABO=AB*r1
OEDQGF = ACDQG+ABEOF+ABC, combine above
ABC*(r1+r2)/r=AB*r1+AC*r2+ABC …… (1)
using standard formula
ABC=(AB+AC+BC)r/2 = (AB+AC+BT+CT)r/2 …… (2)
in addition,
AB-BT=AB-BR=AR=AS=AC-SC=AC-CT …(3)
solve system of equations (1), (2), and (3), we get
ABC=BT*r2+CT*r1