Se7en

Let $\triangle A'B'C'$ be the pedal (orthic) triangle of an acute $\triangle ABC$ with an orthocenter $H$ .

Since $\angle HA'B = \angle HC'B = \frac{\pi}{2}$ from the altitudes of $\triangle ABC$ , $BC'HA'$ is a cyclic quadrilateral, and since they subtend the same arc, $\angle HBA' = \angle HC'A'$ . Also, $\angle BB'C = \frac{\pi}{2}$ from an altitude of $\triangle ABC$ , so from $\triangle BB'C$ , $\angle HBA' = \frac{\pi}{2} - \angle BCB'$ or $\angle HC'A' = \frac{\pi}{2} - \angle C$ . Similarly, $\angle HC'B' = \frac{\pi}{2} - \angle C$ , so $\angle C' = \angle HC'A' + \angle HC'B' = \pi - 2 \angle C$ . By similar arguments, $A' = \pi - 2A$ , $B' = \pi - 2B$ , and $C' = \pi - 2C$ .

Without loss of generality, if $\triangle A'B'C' \sim \triangle ABC$ , then either:

1) $A = \pi - 2A$ , $B = \pi - 2B$ , and $C = \pi - 2C$ : which solves to $A = B = C = \frac{\pi}{3}$ , an equilateral triangle

2) $A = \pi - 2B$ , $B = \pi - 2A$ , and $C = \pi - 2C$ : which solves to $A = B = C = \frac{\pi}{3}$ , an equilateral triangle

3) $A = \pi - 2B$ , $B = \pi - 2C$ , and $C = \pi - 2A$ : which solves to $A = B = C = \frac{\pi}{3}$ , an equilateral triangle

In all cases, an acute triangle that is similar to its own pedal triangle must be an equilateral triangle. Since a pedal triangle does not exist for a right triangle, a non-equilateral triangle $\triangle ABC$ similar to its own pedal triangle must be obtuse. Without loss of generality, let $\angle A$ be obtuse.

Then by the same reasoning from above, $\angle B'C'A' = \pi - 2\angle BCB'$ . But since $\angle BB'C = \frac{\pi}{2}$ from an altitude of $\triangle ABC$ , then from $\triangle BB'C$ we have $\angle BCB' = \frac{\pi}{2} - \angle CBA$ . By substitution, $\angle B'C'A' = \pi - 2(\frac{\pi}{2} - \angle CBA) = 2\angle CBA$ . Therefore, $C' = 2B$ and similarly $B' = 2C$ . This means that $A' = \pi - B' - C' = \pi - 2C - 2B = \pi - 2(B + C) = \pi - 2(\pi - A) = 2A - \pi$ .

If $\triangle A'B'C' \sim \triangle ABC$ with $\angle A$ being obtuse, then either $A = 2A - \pi$ or $A = 2B$ . If $A = 2A - \pi$ then $A = \pi$ , which is not possible in a triangle, so $A = 2B$ . Then either $C = 2C$ or $C = 2A - \pi$ . If $C = 2C$ then $C = 0$ , which is not possible in a triangle, so $C = 2A - \pi$ , which means $B = 2C$ . Solving $A = 2B$ , $B = 2C$ , and $C = 2A - \pi$ leads to $A = \frac{4\pi}{7}$ , $B = \frac{2\pi}{7}$ , and $C = \frac{\pi}{7}$ .

Let $BC = k \sin \frac{4\pi}{7}$ for some value of $k$ . Then by the law of sines, $AC = k \sin \frac{2\pi}{7}$ and $AB = k \sin \frac{\pi}{7}$ , by right angle trigonometry the altitude through $B$ is $h_2 = k \sin \frac{\pi}{7} \sin \frac{4\pi}{7}$ , and by the law of sines the internal bisector through $C$ is $b_1 = \frac{k \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}}{\sin (\pi - \frac{\pi}{14} - \frac{2\pi}{7})} = \frac{k \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}}{\sin \frac{9\pi}{14}}$ .

By the sine product to sum formula, $\sin \frac{\pi}{7} \sin \frac{9\pi}{14} = \frac{1}{2} \cos \frac{3\pi}{14} = \frac{1}{2} \sin \frac{4\pi}{14}$ , so $2h_2 = 2k \sin \frac{\pi}{7} \sin \frac{4\pi}{7} = \frac{k \sin \frac{2\pi}{7} \sin \frac{4\pi}{7}}{\sin \frac{9\pi}{14}} = b_1$ , which means $\boxed{b_1 = 2h_2}$ for a non-equilateral triangle that is similar to its own pedal triangle.