Let $ABC$ be a triangle with $AB=c$, $BC=a$, and $CA=b$. Let $D$, $E$ and $F$ denote the feet of the altitudes from $A$, $B$ and $C$ receptively such that $AD=h_a$, $BE=h_b$ and $CF=h_c$.
First off, we define some notations. Let $L$ be the length of the line drawn by selecting a random point $P$ in a triangle $ABC$ and let $X$ be the point where the random line intersects the triangle.
Let $\mathbb{E}_{\triangle ABC}(L|S_k)$ be the expected length of the line drawn as described in setup $S_k$ in the triangle $ABC$.
Setup 1 Select a random point in triangle $ABC$ and draw a line (towards $BC$) parallel to the altitude $AD$.
WLOG, we can chose the vertex $B$ to be the origin with the $x$-axis aligned with the side $BC$. Then we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_1)=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}|PX|\,dy\,dx=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}y\,dy\,dx=\frac{h_a}{3}$
The last equality follows by recognizing the integral as the $y$-coordinate of the centroid of the triangle $ABC$.
Setup 2 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (towards $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
Using the same co-ordinate system as above, this time we have,
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L|S_2,M)&=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}|PX|\,dy\,dx\\
&=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}\frac{y}{\cos t}\,dy\,dx=\frac{h_a}{3\cos t}=\frac{|AM|}{3}
\end{align}
$
Setup 3 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (away from $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
Now we do a case distinction based on whether the random point is in $\triangle ABM$ or $\triangle AMC$.
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L|S_3,M)&=\frac{\triangle MAB}{\triangle ABC}\mathbb{E}_{\triangle MAB}(L|S_2,A)+\frac{\triangle MCA}{\triangle ABC}\mathbb{E}_{\triangle MCA}(L|S_2,A)\\
&=\frac{\triangle ABM}{\triangle ABC}\frac{|MA|}{3}+\frac{\triangle AMC}{\triangle ABC}\frac{|MA|}{3}\\
&=\frac{|AM|}{3}\\
\end{align}
$
Surprisingly, it doesn't matter whether we draw the line towards or away from the edge $BC$, the expected value remains the same.
Setup 4 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
From the previous discussion, we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_4,M)=\frac{|AM|}{3}$
This can be equivalently written as,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_4,t)=\frac{h_a}{3\cos t}$
Setup 5 Select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) until it meets the triangle, such that the angle $t$ between $PX$ and $AD$ (measured clockwise from $PX$) satisfies $0 \leq t \leq \tan^{-1}\left(\frac{BD}{AD}\right)$.
Using the law of the total probability,
$
\begin{align}
\mathbb{E}_{\triangle ABC}(L|S_5)&=\mathbb{E}(\mathbb{E}_{\triangle ABC}(L|S_4,t))\\
&=\frac{1}{\tan^{-1}\left(\frac{BD}{AD}\right)}\int\limits_0^{\tan^{-1}\left(\frac{BD}{AD}\right)}\frac{h_a}{3\cos t}\,dt\\
\end{align}
$
We can use the fact that $\int\limits_0^{\tan^{-1}m}\sec t\,dt=\text{sinh}^{-1}m$, we have
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_5)=\frac{h_a}{3\tan^{-1}\left(\frac{BD}{AD}\right)}\text{sinh}^{-1}\left(\frac{BD}{AD}\right)$
Setup 6 Select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) until it meets the triangle, such that the angle $t$ between $PX$ and the side $AB$ (measured clockwise from $PX$) satisfies $0 \leq t \leq \angle A$.
We again make a case distinction based on where the ray $PX$ intersects the side $BC$ - either in $BD$ or in $DC$. We can separate the angle into two cases based on this. Using the previous result, we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_6)=\frac{h_a}{3\angle A}\left(\text{sinh}^{-1}\left(\frac{BD}{AD}\right)+\text{sinh}^{-1}\left(\frac{DC}{AD}\right)\right)$
This, after some basic simplifications, results in,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_6)=\frac{h_a}{3\angle A}\text{csch}^{-1}\left(\frac{p}{a}\frac{s-a}{p-a}\right)$
where $p$ and $s$ are the perimeter and semi-perimeter of $\triangle ABC$ respectively.
Now we get into the final part of the proof.
Setup 7 Select a random point in triangle $ABC$ and draw a line at a random angle from that point until it meets the triangle.
We now again do a case distinction based on the angle and using the previous result, we finally have,
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L)&=\sum_{x \in \{a,b,c\}}\frac{h_x}{3\pi}\text{csch}^{-1}\left(\frac{p}{x}\frac{s-x}{p-x}\right)\\
&=\frac{h_a}{3\pi}\text{csch}^{-1}\left(\frac{p}{a}\frac{s-a}{p-a}\right)+\frac{h_b}{3\pi}\text{csch}^{-1}\left(\frac{p}{b}\frac{s-b}{p-b}\right)+\frac{h_c}{3\pi}\text{csch}^{-1}\left(\frac{p}{c}\frac{s-c}{p-c}\right)
\end{align}
$
And hence the result. Hope you enjoyed this. I'll come up with something different the next time.
First off, we define some notations. Let $L$ be the length of the line drawn by selecting a random point $P$ in a triangle $ABC$ and let $X$ be the point where the random line intersects the triangle.
Let $\mathbb{E}_{\triangle ABC}(L|S_k)$ be the expected length of the line drawn as described in setup $S_k$ in the triangle $ABC$.
Setup 1 Select a random point in triangle $ABC$ and draw a line (towards $BC$) parallel to the altitude $AD$.
WLOG, we can chose the vertex $B$ to be the origin with the $x$-axis aligned with the side $BC$. Then we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_1)=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}|PX|\,dy\,dx=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}y\,dy\,dx=\frac{h_a}{3}$
The last equality follows by recognizing the integral as the $y$-coordinate of the centroid of the triangle $ABC$.
Setup 2 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (towards $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
Using the same co-ordinate system as above, this time we have,
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L|S_2,M)&=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}|PX|\,dy\,dx\\
&=\frac{1}{\triangle ABC}\iint\limits_{(x,y)\in\triangle ABC}\frac{y}{\cos t}\,dy\,dx=\frac{h_a}{3\cos t}=\frac{|AM|}{3}
\end{align}
$
Setup 3 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (away from $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
Now we do a case distinction based on whether the random point is in $\triangle ABM$ or $\triangle AMC$.
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L|S_3,M)&=\frac{\triangle MAB}{\triangle ABC}\mathbb{E}_{\triangle MAB}(L|S_2,A)+\frac{\triangle MCA}{\triangle ABC}\mathbb{E}_{\triangle MCA}(L|S_2,A)\\
&=\frac{\triangle ABM}{\triangle ABC}\frac{|MA|}{3}+\frac{\triangle AMC}{\triangle ABC}\frac{|MA|}{3}\\
&=\frac{|AM|}{3}\\
\end{align}
$
Surprisingly, it doesn't matter whether we draw the line towards or away from the edge $BC$, the expected value remains the same.
Setup 4 Given a point $M$ between $B$ and $D$, select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) parallel to $AM$ until it meets the triangle. Let $t$ be the angle between $AM$ and $AD$ (measured clockwise from $AD$).
From the previous discussion, we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_4,M)=\frac{|AM|}{3}$
This can be equivalently written as,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_4,t)=\frac{h_a}{3\cos t}$
Setup 5 Select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) until it meets the triangle, such that the angle $t$ between $PX$ and $AD$ (measured clockwise from $PX$) satisfies $0 \leq t \leq \tan^{-1}\left(\frac{BD}{AD}\right)$.
Using the law of the total probability,
$
\begin{align}
\mathbb{E}_{\triangle ABC}(L|S_5)&=\mathbb{E}(\mathbb{E}_{\triangle ABC}(L|S_4,t))\\
&=\frac{1}{\tan^{-1}\left(\frac{BD}{AD}\right)}\int\limits_0^{\tan^{-1}\left(\frac{BD}{AD}\right)}\frac{h_a}{3\cos t}\,dt\\
\end{align}
$
We can use the fact that $\int\limits_0^{\tan^{-1}m}\sec t\,dt=\text{sinh}^{-1}m$, we have
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_5)=\frac{h_a}{3\tan^{-1}\left(\frac{BD}{AD}\right)}\text{sinh}^{-1}\left(\frac{BD}{AD}\right)$
Setup 6 Select a random point in triangle $ABC$ and draw a line (either towards or away from $BC$) until it meets the triangle, such that the angle $t$ between $PX$ and the side $AB$ (measured clockwise from $PX$) satisfies $0 \leq t \leq \angle A$.
We again make a case distinction based on where the ray $PX$ intersects the side $BC$ - either in $BD$ or in $DC$. We can separate the angle into two cases based on this. Using the previous result, we have,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_6)=\frac{h_a}{3\angle A}\left(\text{sinh}^{-1}\left(\frac{BD}{AD}\right)+\text{sinh}^{-1}\left(\frac{DC}{AD}\right)\right)$
This, after some basic simplifications, results in,
$\displaystyle\mathbb{E}_{\triangle ABC}(L|S_6)=\frac{h_a}{3\angle A}\text{csch}^{-1}\left(\frac{p}{a}\frac{s-a}{p-a}\right)$
where $p$ and $s$ are the perimeter and semi-perimeter of $\triangle ABC$ respectively.
Now we get into the final part of the proof.
Setup 7 Select a random point in triangle $ABC$ and draw a line at a random angle from that point until it meets the triangle.
We now again do a case distinction based on the angle and using the previous result, we finally have,
$
\begin{align}
\displaystyle\mathbb{E}_{\triangle ABC}(L)&=\sum_{x \in \{a,b,c\}}\frac{h_x}{3\pi}\text{csch}^{-1}\left(\frac{p}{x}\frac{s-x}{p-x}\right)\\
&=\frac{h_a}{3\pi}\text{csch}^{-1}\left(\frac{p}{a}\frac{s-a}{p-a}\right)+\frac{h_b}{3\pi}\text{csch}^{-1}\left(\frac{p}{b}\frac{s-b}{p-b}\right)+\frac{h_c}{3\pi}\text{csch}^{-1}\left(\frac{p}{c}\frac{s-c}{p-c}\right)
\end{align}
$
And hence the result. Hope you enjoyed this. I'll come up with something different the next time.
UPDATE (13 / 8 / 2020): Updating the 'boring calculation' used in Setup 6.
By the addition formulae for Inverse Hyperbolic sines,
$\displaystyle \text{sinh}^{-1}\left(\frac{BD}{AD}\right)+\text{sinh}^{-1}\left(\frac{CD}{AD}\right)=\text{sinh}^{-1}\left(\frac{CD}{AD}\frac{AB}{AD}+\frac{BD}{AD}\frac{AC}{AD}\right)$
Using the geometry of the triangle we know that,
$BD=c \cos B$, $CD=b \cos C$ and $AD=b \sin C=c \sin B$
$\displaystyle \text{sinh}^{-1}\left(\frac{BD}{AD}\right)+\text{sinh}^{-1}\left(\frac{CD}{AD}\right)=\text{sinh}^{-1}\left(\frac{\cos B+\cos C}{\sin B \sin C}\right)$
But
$\sin B \sin C=2\sin(B/2)\cos(B/2)\cdot 2\sin(C/2)\cos(C/2)=2\sin(B/2)\sin(C/2)\cdot 2\cos(B/2)\cos(C/2)$
With Sine and Cosine addition formula, this becomes,
$\sin B \sin C=(\cos(B/2+C/2)-\cos(B/2-C/2))(\cos(B/2+C/2)+\cos(B/2-C/2))=\cos^2(B/2+C/2)-\cos^2(B/2-C/2)$
Also, $\displaystyle \cos B+\cos C=2\cos(B/2+C/2)\cos(B/2-C/2)$
Therefore,
$\displaystyle \frac{\sin B \sin C}{\cos B+\cos C}=\frac{1}{2}\left( \frac{\cos(B/2+C/2)}{\cos(B/2-C/2)}-\frac{\cos(B/2-C/2)}{\cos(B/2+C/2)} \right)$
The rest follows by Mollweide's formula.
Until then,
Yours Aye
Me
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