The Simson line

Idea. A complex-number computation avoids every configuration issue.

Step 1: the foot of the perpendicular. Let $z$ be the foot of the perpendicular from $p$ to the line through $b,c$ (with $|b|=|c|=1$, $b\ne c$).
- $z$ lies on line $bc$: $z+bc\,\bar z=b+c$.
- $z-p$ is perpendicular to $b-c$: $\frac{z-p}{b-c}+\overline{\left(\frac{z-p}{b-c}\right)}=0$, i.e. $z-p=bc(\bar z-\bar p)$.

Substituting $bc\bar z=b+c-z$: $z-p=b+c-z-bc\bar p$, so
$$z=\frac{p+b+c-bc\,\bar p}{2}=\frac{p+b+c-\frac{bc}{p}}2.$$

Step 2: differences of feet. Let $f_a,f_b,f_c$ be the feet on $BC,CA,AB$:
$$f_a=\frac{p+b+c-bc/p}2,\qquad f_b=\frac{p+c+a-ca/p}2,\qquad f_c=\frac{p+a+b-ab/p}2.$$
Then
$$f_a-f_b=\frac{(b-a)(p-c)}{2p},\qquad f_a-f_c=\frac{(c-a)(p-b)}{2p}.$$
So
$$\frac{f_a-f_b}{f_a-f_c}=\frac{(b-a)(p-c)}{(c-a)(p-b)}=:Q.$$

Step 3: $Q$ is real. Use $\overline{x-y}=\frac{y-x}{xy}$ for unit numbers:
$$\bar Q=\frac{(a-b)(c-p)}{(a-c)(b-p)}\cdot\frac{ac\cdot bp}{ab\cdot cp}=\frac{(a-b)(c-p)}{(a-c)(b-p)}=Q.$$
Thus $Q\in\mathbb R$, which means that $f_a,f_b,f_c$ are collinear. $\blacksquare$

Remark. The converse also holds. The Simson line of $P$ bisects the segment joining $P$ to the orthocentre $H$.

Lesson. On the unit circle, feet of perpendiculars have the neat formula $\frac{p+b+c-bc\bar p}2$; collinearity becomes "a quotient is real".