Denote by p1 the semiperimeter of ABK, and by p2 the semiperimeter of AKC. We have to prove that (p1-AK)(p2-AK)>(p1-c)(p2-b) after opening the paranthesis the conclusion follows.

Sailor wrote: Denote by p1 the semiperimeter of ABK, and by p2 the semiperimeter of AKC. We have to prove that (p1-AK)(p2-AK)>(p1-c)(p2-b) after opening the paranthesis the conclusion follows. Hmm, that took me some time to understand... In order to make your solution a bit clearer for beginners, I will write a few more details: We have $BM=\frac{AB+BK-AK}{2}$, $CN=\frac{AC+CK-AK}{2}$, $KM=\frac{AK+BK-AB}{2}$, $KN=\frac{AK+CK-AC}{2}$; thus, proving the inequality $BM\cdot CN > KM\cdot KN$ reduces to proving $\left(AB+BK-AK\right)\cdot\left(AC+CK-AK\right)>\left(AK+BK-AB\right)\cdot\left(AK+CK-AC\right)$. Now, after opening the brackets, this simplfies to $AC\cdot BK + AB\cdot CK > AK\cdot BK+AK\cdot CK$, or, equivalently, $AC\cdot BK + AB\cdot CK > AK\cdot BC$. But this follows from the Ptolemy inequality, applied to the quadrilateral ABKC (equality cannot occur since the quadrilateral ABKC can never be cyclic). Anyway, here is my solution of the problem: Let $s=\frac{a+b+c}{2}$ be the semiperimeter of triangle ABC. Let r be the inradius and $r_a$, $r_b$, $r_c$ be the exradii of triangle ABC. Then, it is well-known that $\cot\frac{B}{2}=\frac{r_c}{s-a}$ and $\cot\frac{C}{2}=\frac{s}{r_c}$; thus, $\cot\frac{B}{2}\cdot\cot\frac{C}{2}=\frac{r_c}{s-a}\cdot\frac{s}{r_c}=\frac{s}{s-a}$. Since, obviously, s > s - a, we have $\frac{s}{s-a}>1$, and thus, $\cot\frac{B}{2}\cdot\cot\frac{C}{2}>1$. Now, the point M was defined as the point where the incircle of triangle ABK touches the line BC. Thus, if X is the center of this incircle, we have $XM\perp BC$, and consequently, the triangles BMX and KMX are right-angled. It follows that ${BM=XM\cdot\cot\measuredangle XBM}$ and $KM=XM\cdot\cot\measuredangle XKM$. Thus, $\frac{BM}{KM}=\frac{\cot\measuredangle XBM}{\cot\measuredangle XKM}$. But since the point X is the center of the incircle of triangle ABK, the lines KX and BX are the angle bisectors of the angles BKA and ABK. Thus, $\measuredangle XKM=\frac{\measuredangle BKA}{2}$ and $\measuredangle XBM=\frac{\measuredangle ABK}{2}$. The latter of these equations rewrites as $\measuredangle XBM=\frac{B}{2}$. Thus, $\frac{BM}{KM}=\frac{\cot\measuredangle XBM}{\cot\measuredangle XKM}=\frac{\cot\frac{B}{2}}{\cot\frac{\measuredangle BKA}{2}}$. Similarly, $\frac{CN}{KN}=\frac{\cot\frac{C}{2}}{\cot\frac{\measuredangle CKA}{2}}$. Hence, $\frac{BM\cdot CN}{KM\cdot KN}=\frac{BM}{KM}\cdot\frac{CN}{KN}=\frac{\cot\frac{B}{2}}{\cot\frac{\measuredangle BKA}{2}}\cdot\frac{\cot\frac{C}{2}}{\cot\frac{\measuredangle CKA}{2}}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}}$. Since < BKA + < CKA = 180°, we have $\frac{\measuredangle BKA}{2}+\frac{\measuredangle CKA}{2}=\frac{\measuredangle BKA+\measuredangle CKA}{2}=\frac{180^{\circ}}{2}=90^{\circ}$. Hence, $\cot\frac{\measuredangle BKA}{2}=\tan\frac{\measuredangle CKA}{2}$. Thus, $\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}=\tan\frac{\measuredangle CKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}$ $=\tan\frac{\measuredangle CKA}{2}\cdot\frac{1}{\tan\frac{\measuredangle CKA}{2}}=1$. And consequently, $\frac{BM\cdot CN}{KM\cdot KN}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{\cot\frac{\measuredangle BKA}{2}\cdot\cot\frac{\measuredangle CKA}{2}}=\frac{\cot\frac{B}{2}\cdot\cot\frac{C}{2}}{1}=\cot\frac{B}{2}\cdot\cot\frac{C}{2}$. But we know that $\cot\frac{B}{2}\cdot\cot\frac{C}{2}>1$; thus, $\frac{BM\cdot CN}{KM\cdot KN}>1$, so that $BM\cdot CN>KM\cdot KN$. Proof complete. darij