Let I be the incenter of the triangle $\triangle ABC$. We have to show that the bisector AI of the angle $\angle A$ is perpendicular to EF iff either AB = AC or the angle $\angle CAB = 90^\circ$ is right. The bisectors $AI, BI \equiv EI, CI \equiv FI$ are concurent cevians of the triangle $\triangle AEF$. Let $X \in AF, Y \in AE$ be the feet of EI, FI in this triangle. $\angle EXA \equiv \angle BXA = 180^\circ - \frac{\angle B}{2} - (90^\circ - \angle B) - \frac{90^\circ - \angle C}{2} =$ $= \frac{90^\circ + \angle B + \angle C}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$ $\angle FYA \equiv \angle CYA = 180^\circ - \frac{\angle C}{2} - (90^\circ - \angle C) - \frac{90^\circ - \angle B}{2} =$ $= \frac{90^\circ + \angle C + \angle B}{2} = 90^\circ + \frac{90^\circ - \angle A}{2}$ Hence, $\angle EXA = \angle FYA$. If the angle $\angle A = 90^\circ$ is right, EX, FY are 2 altitudes of the triangle $\triangle AEF$, I its orthocenter and $AI \perp EF$ its remaining altitude. Assume now that the angle $\angle A \neq 90^\circ$ is not right and that we still have $AI \perp EF$, i.e., AI is the A-altitude of the triangle $\triangle AEF$, but I is no longer its orthocenter. Since $\angle EXA = \angle FYA \neq 90^\circ$, then also $\angle EXF = \angle FYE \neq 90^\circ$, i.e., the quadrilateral EFXY is cyclic. Let (P) be its circumcircle, which is centered on the perpendicular bisector of the segment EF. Using parallel projection, we can project the triangle $\triangle AEF$ into a triangle $\triangle A'EF$, so that the point I is projected into its orthocenter I'. Then the cevians EX, FY are projected into its altitudes EX', FY', i.e., the quadrilateral EFX'Y' is also cyclic, with the circumcircle (O') centered at the midpoint O' of the segment EF. Thus the points X, Y lie both on the circle (P) and on an ellipse $o$ with the main axis EF (which is projected into the circle (O') in our parallel projection), not identical with the circle (P). Since both the ellipse $o$ and the circle (P) are symmetrical with respect to the perpendicular bisector of the segment EF, so are their intersections X, Y, i.e. EX = FY and EY = FX. Hence, the triangles $\triangle EFX \cong \triangle FEY$ are congruent and it immediately follows that the triangle $\triangle AEF$ is isosceles with AE = AF. This and $AD \perp BC$ implies that the incircles $(E) \cong (F)$ are congruent, which means that the triangle $\triangle ABC$ itself is isosceles with AB = AC.