Proof First, we will prove that three points $A$, $I$ and $A_1$ are collinear $\Leftrightarrow \angle BAC = 60^\circ$.
As shown in the figure, assume that $O$ is the circumcenter of $\triangle ABC$. We join $BO$ and $CO$, then

$$\angle BHC = 180^\circ - \angle BAC,$$
$$\angle BA_1C = 2(180^\circ - \angle BHC) = 2\angle BAC.$$
$$\text{Hence, } \angle BAC = 60^\circ \Leftrightarrow \angle BAC + \angle BA_1C = 180^\circ$$
$$\Leftrightarrow A_1 \text{ is on the circumcircle } \odot O \text{ of } \triangle ABC$$
$$\Leftrightarrow AI \text{ and } AA_1 \text{ coincide (because } A_1 \text{ is on the perpendicular bisector of } BC. )$$
$$\Leftrightarrow \text{Three points } A, I \text{ and } A_1 \text{ are collinear.}$$

Secondly, we will prove $S_{\triangle BKB_2} = S_{\triangle CKC_2} \Leftrightarrow \angle BAC = 60^\circ$.

Construct $IP \perp AB$ at point $P$, and $IQ \perp AC$ at $Q$, then
$$S_{\triangle AB_1 B_2} = \frac{1}{2} IP \cdot AB_2 + \frac{1}{2} IQ \cdot AB_1.$$
Note that $S_{\triangle AB_1 B_2} = \frac{1}{2} AB_1 \cdot AB_2 \cdot \sin A$,
therefore $IP \cdot AB_2 + IQ \cdot AB_1 = AB_1 \cdot AB_2 \cdot \sin A$.
Assume $IP = r$, where $r$ is the radius of the inscribed circle of $\triangle ABC$. Then $IQ = r$. Furthermore set $BC = a$, $CA = b$ and $AB = c$, then $r = \frac{2S_{\triangle ABC}}{a+b+c}$.
From $AB_1 = \frac{b}{2}$ and $2AB_1 \cdot \sin A = h_c = \frac{2S_{\triangle ABC}}{c}$,
we have $AB_2 \cdot \left[ \frac{2S_{\triangle ABC}}{c} - 2 \cdot \frac{2S_{\triangle ABC}}{a+b+c} \right] = b \cdot \frac{2S_{\triangle ABC}}{a+b+c}$,
so $AB_2 = \frac{bc}{a+b+c}$.
Similarly, $AC_2 = \frac{bc}{a+c-b}$.
Hence,
$$S_{\triangle BKB_2} = S_{\triangle CKC_2}$$
$$\Leftrightarrow S_{\triangle ABC} = S_{\triangle AB_2 C_2}$$
$$\Leftrightarrow bc = \frac{bc}{a+b-c} \cdot \frac{bc}{a+c-b}$$
$$\Leftrightarrow a^2 = b^2 + c^2 - bc$$
$$\Leftrightarrow \angle BAC = 60^\circ \text{ (By the law of cosines).}$$