Please verify and critique my proof of Viviani's Theorem: In an equilateral triangle, the sum of the distances from any point within the triangle to each of its sides equals the altitude of the triangle.
Note: Many proofs are available. This question is to verify, critique, or offer feedback to my proof, below. The usage of the lemma is inspired by Kiselev's Geometry, Vol. 1, Ex. 187.
Lemma: Let $\triangle ABC$ be an arbitrary isosceles triangle with $\angle ABC = \angle ACB$. Let $P$ be an arbitrary point on segment $BC$. Then $d(P, AB) + d(P, AC) = d(B, AC)$.
To show this, drop the perpendicular from $B$ to $AC$ at $H$, from $P$ to $AC$ at $C'$, from $P$ to $AB$ at $B'$, and from $P$ to $BH$ at $Y$. Clearly, $BH$ and $PB'$ intersect at a point within $\triangle ABC$; label it $X$. It suffices to show $PB' + PC' = BH$.
Since $YH$ and $PC'$ are both perpendicular to $AC$, they are parallel, and since $\angle PYH$ is right, then $YHC'P$ is a rectangle, and $YH \cong PC'$. It remains to show that $BY \cong PB'$.
[*] Since $\angle XBP$ and $\angle XPB$ are each complementary to $\angle ACB = \angle ABC$, then $\triangle XBP$ is isosceles, and $BX \cong PX$. Then right triangles $\triangle B'XB$ and $\triangle YXP$ are congruent by AAS, and $XY \cong XB'$, giving $BY \cong PB'$ and proving the lemma.
Main Proof: Let $\triangle ABC$ be an equilateral triangle and point $P$ within it. Draw $B'$ on $AB$ such that $B'P \parallel BC$ and $d(B', BC) = d(P, BC)$. Extend $B'P$ to intersect $AC$ at $C'$. Then $\triangle AB'C'$ is isosceles and $d(P,AB) + d(P, AC) + d(P, BC) = d(B', AC) + d(B', BC)$. But $\triangle CAB$ is also isosceles, so $d(B', AC) + d(B', BC) = d(B, AC).$ QED.
Based on Blue's comment, the paragraph marked [*] can be replaced as follows:
Since $YP$ and $AC$ are both perpendicular to $BH$, they are parallel, and $\angle YPB \cong \angle ACB \cong \angle B'BP$. Likewise, $\angle B'PB$ and $\angle YBP$ are each complementary to the same angle $\angle ACB = \angle ABC$ and therefore congruent. Therefore, $\triangle B'BP \cong \triangle YPB$ by ASA, giving $BY \cong PB'$ and proving the lemma.
Best Answer
The argument is sound, but could use a bit of refinement. Below is how I might present it here on Math.SE. In a more formal setting, I'd likely be more verbose and less prone to fold explanations into the equations themselves ... perhaps to the detriment of comprehensibility.
Proof of Lemma. Drop perpendiculars from $P$ to $B'$ and $C'$ on lines $AB$ and $AC$, respectively; from $B$ to $H$ on line $AC$; and from $P$ to $Y$ on line $BH$. Note that $\square PC'HY$ is a rectangle. Also, $$\angle B'BP \underbrace{\quad\cong\quad}_{\text{isos. $\triangle ABC$}} \angle ACB \underbrace{\quad\cong\quad}_{YP\parallel AC} \angle YPB$$ so that right triangles $\triangle PB'B$ and $\triangle BYP$ are congruent by Hypotenuse-(Acute)Angle. Therefore, we can write
$$\begin{align} d(P,AB) \;+\; d(P,AC) &\;=\; \underbrace{|PB'|} \;+\;\;\; |PC'| \\ &\;=\; \;\;\color{blue}{|BY|}\;\; +\; \underbrace{|PC'|} &(\text{$\triangle PB'B\cong\triangle BYP$}) \\ &\;=\; \;\;|BY|\;\;+\;\;\;\color{red}{|YH|} &(\text{$\square PC'HY$ is a rectangle}) \\[6pt] &\;=\;\;\;|BH| \;=\;d(B,AC) \end{align}$$
Done! $\square$
Proof of Theorem. Consider point $P$ within equilateral $\triangle ABC$. Let the line through $P$ parallel to line $BC$ meet $AB$ and $AC$ at $B'$ and $C'$, respectively. Since both $\triangle AB'C'$ and $\triangle CAB$ are isosceles, we can simplify the sum of the distances from $P$ to the sides of $\triangle ABC$ thusly:
$$\begin{align} \phantom{=}\quad \underbrace{d(P,BC)}\;&+\;d(P,AC)\;+\;d(P,AB) \\ =\quad \color{blue}{d(B',BC)}\;&+\;\underbrace{d(P,AC)\;+\;d(P,AB)} &(\text{$P$ and $B'$ are equidistant from $BC$})\\ =\quad d(B',BC)\;&+\;\qquad \color{red}{d(B',AC)} &(\text{Lemma applied to $P$ in isos. $\triangle AB'C'$})\\[6pt] =\quad\quad\quad\;\; d(A&,BC) &(\text{Lemma applied to $B'$ in isos. $\triangle CAB$}) \end{align}$$ Done! $\square$
The use of underbraces and color may be a bit excessive (the latter also being problematic for color-blind readers), but I think they help, especially in the second string of arguments with all those $d(X,YZ)$ terms that might otherwise become a visual jumble. Even without those enhancements, simply using "displayed" equations makes the changes from one expression to the next easier to scan than using their in-line counterparts.
$$\begin{align} \phantom{=}\quad d(P,BC)\;&+\;d(P,AC)\;+\;d(P,AB) \\[2pt] =\quad d(B',BC)\;&+\;d(P,AC)\;+\;d(P,AB)\\[2pt] =\quad d(B',BC)\;&+\;\qquad d(B',AC) \\[2pt] =\quad\quad\quad\;\; d(A&,BC) \end{align}$$
I'll note that I also incorporated extra space around the "$+$"s and "$=$"s, and some vertical padding, to improve on this rather dense alternative:
$$\begin{align} \phantom{=} d(P,BC)&+d(P,AC)+d(P,AB) \\ =d(B',BC)&+d(P,AC)+d(P,AB) \\ =d(B',BC)&+\qquad d(B',AC) \\ =\quad\quad\; d(A&,BC) \end{align}$$