TITLE: What is so special about triangles?!
QUESTION [17 upvotes]: Take any random triangle. 
If we draw internal-angle-bisectors of all its angles, they intersect at the same point.
 
If we draw the perpendicular bisectors of each side (although they aren't cevians), they are concurrent too.
 
The same goes for the altitudes, and medians corresponding to each vertex
Of course any two non-parallel lines in a plane will intersect, but in triangles, all three lines are concurrent! Is there any special reason behind this? Has it been explained? I couldn't find any reasoning on the net.

So far, I've got proofs, but what I am actually looking for is what @GregHill said:

Does anyone know a way to explain the idea of ALL triangle concurrences based on something intrinsic about triangles?"

Why do triangles have this property of having cevians (and other lines) concurrent?
$P.S:$ Thanks to @Blue for the correct word for the 'lines' of a triangle: Cevians :)

REPLY [6 votes]: Here's an extension of Ceva's theorem that incorporates the concurrency of perpendicular bisectors (and, actually, (almost) any trio of lines).

Let distinct lines $p := \overleftrightarrow{P_1P_2}$, $q := \overleftrightarrow{Q_1Q_2}$, $r := \overleftrightarrow{R_1R_2}$ meet the (extended) edges of non-degenerate $\triangle ABC$ such that
$$P_1 = A + p_1 (B-A) \qquad Q_1 = B + q_1 (C-B) \qquad R_1 = C + r_1(A-C)$$
$$P_2 = A + p_2 (C-A) \qquad Q_2 = B + q_2 (A-B) \qquad R_2 = C + r_2(B-C)$$
for some $p_1$, $p_2$, $q_1$, $q_2$, $r_1$, $r_2$. Then $p$, $q$, $r$ concur if and only if
$$\begin{array}{c}
p_1 q_1 r_1 ( 1 - p_2 - q_2 - r_2 ) + p_2 q_2 r_2 ( 1 - p_1 - q_1 - r_1 ) \\[4pt]
+\; p_1 q_1 \; q_2 r_2 \;+\; q_1 r_1 \; r_2 p_2 \;+\; r_1 p_1 \; p_2 q_2 \quad = \quad 0   
\end{array} \qquad (\star)$$
(Proof ---for instance, using straightforward vector methods--- is left as an exercise for the reader.)

When $p_1 = q_1 = r_1 = 1$, then $p$, $q$, $r$ are cevians through $B$, $C$, $A$, respectively, and $(\star)$ reduces to 
$$( 1 - p_2 )( 1 - q_2 )( 1 - r_2 ) = p_2 q_2 r_2 \qquad (\star\star)$$
which recaptures Ceva's Theorem.

If $p$, $q$, $r$ are perpendicular to $\overline{AB}$ (of length $c$), $\overline{BC}$ (of length $a$), $\overline{CA}$ (of length $b$), respectively, then, via right triangle $\triangle AP_1 P_2$,
$$|AP_1| = |AP_2|\cos A \quad\implies\quad c p_1 = b p_2 \cos A = b p_2 \frac{-a^2+b^2+c^2}{2bc}$$
whence
$$
p_2 = \frac{2 c^2 p_1}{-a^2+b^2+c^2} \qquad
q_2 = \frac{2 a^2 q_1}{a^2-b^2+c^2} \qquad 
r_2 = \frac{2 b^2 r_1}{a^2+b^2-c^2}$$
and (for non-degenerate $\triangle ABC$) condition $(\star)$ reduces to 
$$a^2 + b^2 + c^2 = 2 \left( a^2 q_1 + b^2 r_1 + c^2 p_1 \right) \qquad (\star\star\star)$$
If $p$, $q$, $r$ are, more-specifically, perpendicular bisectors of the triangle's edges, then $p_1 = q_1 = r_1 = 1/2$. This satisfies $(\star\star\star)$, so the lines are concurrent.

Edit. After bouncing this idea around a bit, I hit upon a better rendition of the Ceva extension, including a more-Ceva-like version of $(\star)$. I'll take this opportunity to rename some points.

Extended Ceva's Theorem. Consider points $D_1$, $D_2$, $E_1$, $E_2$, $F_1$, $F_2$ on the (extended) edges of $\triangle ABC$, with $D_i$, $E_i$, $F_i$ on the (extended) edge opposite vertex $A$, $B$, $C$, respectively.

Lines $\overleftrightarrow{D_1E_2}$, $\overleftrightarrow{E_1F_2}$, $\overleftrightarrow{F_1D_2}$ concur if and only if
  $$\begin{align}
1 &= \frac{|BD_1|}{|D_1C|} \frac{|CE_1|}{|E_1A|} \frac{|AF_1|}{|F_1B|} 
+ \frac{|D_2C|}{|BD_2|} \frac{|E_2A|}{|CE_2|} \frac{|F_2B|}{|AF_2|} \\[6pt]
&+ \frac{|BD_1|}{|D_1C|} \frac{|D_2C|}{|BD_2|}
+ \frac{|CE_1|}{|E_1A|} \frac{|E_2A|}{|CE_2|}
+ \frac{|AF_1|}{|F_1B|} \frac{|F_2B|}{|AF_2|} \qquad\qquad (\star\star\star\star)
\end{align}$$
  Note: The above uses signed lengths, with $\overrightarrow{AB}$, $\overrightarrow{BC}$, $\overrightarrow{CA}$ indicating the direction of a positively-signed segment on each of the triangle's sides. 

We get Ceva's Theorem back from the Extended Ceva's Theorem by moving $D_2$, $E_2$, $F_2$ to coincide with $C$, $A$, $B$, respectively, so that $|D_2C| = |E_2A| = |F_2B| = 0$; this eliminates all but the first term of the right-hand side of $(\star\star\star\star)$.
For the perpendicular variant, with the three lines ("orthians"?) perpendicular to the sides of $\triangle ABC$ at $D_1$, $E_1$, $F_1$, we can write
$$|D_2C| = |BC| - |BD_2| \qquad
|E_2A| = |CA| - |CE_2| \qquad 
|F_2B| = |AB| - |AF_2|$$
and then also
$$|BD_2| = -\frac{|F_1B|}{\cos B} \qquad
|CE_2| = -\frac{|D_1C|}{\cos C}\qquad
|AF_2| = -\frac{|E_1A|}{\cos A}$$
(the negatives maintain the relationships of the signed lengths). Expressing the cosines in terms of the lengths of the triangles edges, and then expressing those lengths as
$$|BC| = |BD_1|+|D_1C| \qquad |CA| = |CE_1| + |E_1C| \qquad |AB| = |AF_1|+|F_1B|$$ 
equation $(\star\star\star\star)$ eventually reduces to something much nicer than $(\star\star\star)$; namely,
$$|BD_1|^2 + |CE_1|^2 + |AF_1|^2 = |D_1C|^2 + |E_1A|^2 + |F_1B|^2$$
(In a recent answer, I call the above result "Ortha's Theorem" and provide a stand-alone geometric proof.)