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Let us select the points $X$ and $Y$ on the side $BC$ such that $BX = AX$ and $CY = YA$. Then $\angle AXY = \angle ABC + \angle BAX = 2\angle ABC = 60^\circ$ (fig. 39).

Analogously, $\angle XYA = 60^\circ$ and then $\triangle XYA$ is equilateral. Then $BX = AX = XY = AY = YC$, i.e. the points $X$, $Y$ divide $BC$ into three equal parts and $BC = 3AX$.

Let $M$ be the midpoint of $BC$. $\triangle ABC$ is an isosceles triangle, therefore $\angle AMB = 90^\circ$.

Consider the points $B_1, T, K, N, C_1$, where $B$ is the middle of $AB_1$, $X$ is the middle of $AT$, $M$ is the middle of $AN$, $C$ is the middle of $AC_1$. By Thales' theorem, the points $T, K, N$ lie on the segment $B_1C_1$.

We have that $BX = AX = XT$, and therefore $\angle ABT = 90^\circ$. Obviously, $\angle TNA = 90^\circ$. Thus, the quadrangle $ABTN$ is inscribed in a circle $w$. Hence, $\angle ANB = \angle ATB = 90^\circ - \angle BAT = 60^\circ$.

Point $K$ lies on the segment $B_1C_1$, i.e. the points $T, K, N$ lie in one half-plane with respect to the line $AB$. Then, as the arc $\cup BTA$ of the circle $w$ has length $60^\circ$, then from $\angle AKB > 60^\circ$ it follows that the point $K$ lies inside the circle $w$.

On the other hand, the point $K$ lies on the segment $B_1C_1$, hence $K$ lies inside the segment $TN$. Then $D$ lies inside the segment $XM$. We obtain that $AM \perp XM$ and the point $D$ is closer to $M$ than to $X$. Then $AD < AX$, and therefore $3AD < 3AX = BC$, Q.E.D.