Prove the Magic of Morley's Trisector Theorem

 Here's a challenge for the brave geometers out there:

• Take any triangle (doesn't matter how weird its shape is). 
• Trisect all three angles.
• Extends these Trisectors and find where they meet nside the triange.

The Claim: 

The points where these trisectors meet will form a perfect equilateral triangle. Yes, always.

Your Task:

Prove what this magical equilateral triangle always appears. Use any tools or methods you like - geometry, algebra, or even black magic (if that helps). But remember, just saying "It works" doesn't count as proof!

 

Bonus Question: Why do we even bother proving things that fee like geometry is showing off?

Proof

Let $PQR$ be an equilateral triangle. Denote the angle $PQR$ by $pQr$, etc. Choose $L$ outside the triangle $PQR$ such that $$lQr = lRq = \beta + \gamma$$ Clearly $LP$ bisects $qLr$. $M$ and $N$ are similarly defined. Let $LQ$ meet $MP$ in $C$. Then, $$qCp = lQp + mPq - \pi = \alpha + \beta + 2\gamma - \frac{\pi}{3} = \gamma$$ Since $\alpha + \beta + \gamma = \frac{\pi}{3}$ Choose $B$ on $LR$ so that $CP$ bisects $qCb$, $P$ is the in-centre of triangle $LCB$ and so, $$pBr = pBc = \frac{1}{2}(\pi - rLq - lCb) = \beta$$ Giving $bPr + rPn = \pi$. Hence $B$ is the intersection of $LR$ and $NP$. Similarly, if $A$ is the intersection of $MR$ and $NQ$, we have, $$qCa = \gamma; \ qAr = qAc = \alpha$$ $$\text{also,} \ rBa = \beta; \ rAb = \alpha$$ And so $ABC$ is the required triangle, and the theorem is proved.

Reference: Roger Penrose, Collected Works, Volume-1.