What is so special about triangles?!

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 :)

Because the chords given by the edges of a triangle in the circumscribed circle all belong to the same circle, the perpendicular bisector of those chords must go through the center of that circle, and hence all three lines will intersect in the center.

It sounds like you are looking for a single underlying explanation for the concurrence of 3 sets of cevians (angular bisectors, altitudes, and medians), plus perpendicular bisectors, which are not cevians because they do not connect a vertex to the opposite side (unless the triangle is isosceles). These are not the only concurrences associated with triangles. For example the external bisectors of any two angles of a triangle are concurrent with the internal bisector of the third angle. If you aren't looking for a single underlying explanation, you are really asking several different questions to which you will get several different answers!

The 3 cevian concurrences can all be proved using Ceva's theorem, shown in the diagram below, although, for internal bisectors, a proof using Ceva's theorem would be more complicated (outline of easier way is: since any point on any bisector is equidistant from the adjacent sides, and since any two bisectors share one adjacent side, the point where they intersect is equidistant from all three sides). Below the diagram for Ceva's theorem is a diagram outlining how you could use Ceva's to show concurrence of altitudes.

I think the 'single explanation' though would be interesting. Does anyone know a way to explain the idea of ALL triangle concurrences based on something intrinsic about triangles?

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.)

Proof of Ceva's Theorem

We use barycentric coordinates. Since $P_a$ lies on $BC$, the point $P_a$ has the form $P_a=(0,d,1-d)$. So the equation of line $AP_a$ is simply $$ z = \dfrac {1-d}{d} y. $$ Similary, if we let $ P_b = (1 - e, 0, e) $ and $ P_c = (f, 1 - f, 0) $, then the lines $BE$ and $CF$ have equations $ x = \frac {1-e}{e} z $ and $ y = \frac {1 - f}{f} x $, respectively.

Notice that this system of three equations is homogeneous, so we may ignore the barycentric condition that $x+y+z=1$ temporarily. Then, it is easy to see that this equation has solutions if and only if $$ \dfrac {(1-d)(1-e)(1-f)}{def} = 1, $$which is equivalent to Ceva's Theorem.

$ \blacksquare $