Let $N$ be the midpoint of $AC$. Since $K$ is the midpoint of polygonal chain $BAC$, the following holds (Fig. 15):
$$\frac{1}{2}(AB + AC) = KC = KN + NC = KN + \frac{1}{2}AC \Rightarrow KN = \frac{1}{2}AB = NM.$$
By the cosine theorem for $\triangle KNM$:
$$\begin{aligned} & KM^2 = KN^2 + NM^2 - 2 \cdot KN \cdot NM \cdot \cos \angle KNM \\ & \Rightarrow KM^2 = 2KN^2(1 - \cos \angle KNM) = \frac{1}{2} AB^2 (1 + \cos \angle CNM) \end{aligned}$$
Since $\triangle ABC$ is acute angled, $\angle CNM = \angle CAB < 90^\circ \Rightarrow KM^2 > \frac{1}{2} AB^2$, that completes the proof.