Firstly, it is proved that the four triangles of BMK, LNC, IKL and BIC are similar (all three angles are B/2, C/2, 90+A/2).

Then r = KL/[2sin (lik/2)] = KL/[2sin (90+a/2)] = KL/[2cos (a/2)]

R=MN/[2cosIMN]=MN/[2cos(A/2)]

To tangent, you need R=2r.

So MN=2KL

MN=2R/cos(A/2)，KL=R/cos(A/2)

Let a=KL/BC

BC = BM+CN = Rcot(B/2)+Rcot(C/2)= Rcos(A/2)/[sin(B/2)sin(C/2)]

a=sin(b/2)sin(c/2)/[cos(a/2)^2]

The first relationship: 3KL=MK+LN (excited at the sight of 3)

= BMS in(B/2)/sin(C/2)+cns in(C/2)/sin(B/2)

Substitute KL, BM and CN, and simplify it a little.

(sin B+sinC)/[2 sin(B/2)sin(C/2)]= 3/cos(A/2)

Because sinb+sinc = 2 sin (b/2+c/2) cos (b/2-c/2) = 2 cos (a/2) cos (b/2-c/2).

So cos (b/2-c/2) = 3sin (b/2) sin (c/2)/[cos (a/2) 2] = 3a.

The second relationship: MK/CN=BK/CL.

MK=BMsin(B/2)/sin(C/2)

BK = BI+aCI BI = R/sin(B/2)CI = R/sin(C/2)

CL=CI+aBI

The substitution is simplified as1/[2sinb]+asin (c/2) cos (b/2) =1[2sinc]+asin (b/2) cos (c/2).

1/2 *(sin B-sinC)= sin(B/2-C/2)cos(B/2+C/2)= sin(B/2-C/2)sin(A/2)= asin(B/2-C/2)

So sin (a/2) = a.

Finally LK=aBC

MN=2aBC=2sin(A/2)*BC

MN=2AM*sin(A/2)。

So AM=BC

So AB+AC=2AM+BC=3BC.