Because of the fact that $KL=LM$ and that $\angle KLN=\angle NLM=\frac{A}{2}$ we get $S_{AKN}=S_{AMN}$,so the needed identity is $2\cdot KL\cdot AN\cdot cos\frac{A}{2}=KL\cdot (AB+AC)$.Using the sine theorem we trasform it into this:$sin(A+B)+sin(B)=sin(\frac{A}{2}+B)\cdot cos\frac{A}{2}$.Afther opening the brackets we can get: $1+cosA=2cos^2\frac{A}{2}$ which is obiously true. So we are done.

Proof. $\triangle ABN\sim\triangle ALC$ $\implies$ $AN\cdot AL=bc\ (*)$ . Therefore, $KM=AL\cdot \sin A\ \ \wedge\ \ AN\perp KM$ $\implies$ $[AKNM]=\frac {AN\cdot KM}{2}\ \stackrel{(*)}{=}\ \frac {bc\cdot\sin A}{2}=[ABC]$ . An easy extension. In an acute triangle $ABC$ with $b>c$ consider a point $D\in (AC)$ so that $AD=c$ . For a point $L\in (BC)$ denote its projections $K$ , $M$ on $AB$ , $AC$ respectively and the second intersection $N$ of the line $AL$ with the circumcircle of $\triangle CDL$ . Prove that $[AKNM]=[ABC]\cdot\sin \widehat {(AL,KM)}$ .