, 2010

We first show that point $K$ is always inside triangle $\triangle ABC$. Consider the case, when $P = P_1$ lies on the side $AB$ and compute some angles. Let $\angle BAC = \alpha$ (Fig.13), then $\angle BP_1C = 2\alpha$, $\angle ABC = \frac{\pi}{2} - \frac{\alpha}{2}$, $\angle BCP_1 = \frac{\pi}{2} - \frac{3\alpha}{2} \Rightarrow \angle P_1CA = \alpha$, hence, $\triangle ACP_1$ is an isosceles triangle, thus the altitude $P_1K_1$ is also a bisector, therefore $K = K_1$ belongs to the side $AC$. Therefore, as long as $P$ is inside $\triangle ABC$ the bisector $PK$ of adjacent angle of $\angle BPC$ forms the least angle with line $BC$. Thus, perpendicular from $A$ to this line lies between sides $AB$ and $AC$ of triangle $ABC$. Let us extend $AK$ to the intersection with $CP$ (Fig.14) (without loss of generality, we can assume that $P$ is closer to $B$ than to $C$).

Fig. 14

In the same way, $BP$ meets $AK$ at point $X$. Then, in $\triangle PXY$ segment $PK$ is a bisector and an altitude, thus, it is isosceles or $\angle PXY = \angle PYX$, from which it follows that $\angle BXA = \angle AYC$. If we now denote $x = \angle XBA$, $y = \angle YCA$, $x_1 = \angle CAY$, $y_1 = \angle BAX$, then $x_1 + y_1 = \alpha$, $\pi - \alpha = x + y + \angle PBC + \angle PCB = x + y + \pi - 2\alpha \Rightarrow x + y = \alpha = x_1 + y_1$, moreover, $x + y_1 = y + x_1$. From last two identities, we get $x = x_1$, $y = y_1$. Hence, $\triangle ABX = \triangle CAY$ (they have two equal sides and angle between them).

We have the following equalities: $2AK = AX + AY = CY + BX = CY + BP + PX = CY + YP + BP = CP + BP$, which is exactly what we wanted to prove.