Take any random triangle.

If we draw internal-angle-bisectors of all its angles, they intersect at the same point. enter image description here

If we draw the perpendicular bisectors of each side (although they aren't cevians), they are concurrent too.
enter image description here

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 :)

  • $\begingroup$ Another word for those "lines" of a triangle is "cevians", named in honor of the mathematician Ceva who formulated an elegant concurrency condition. See Wikipedia's "Ceva's Theorem" entry. $\endgroup$
    – Blue
    Oct 19, 2013 at 10:41
  • $\begingroup$ Note: perpendicular bisectors are not cevians because they connect side to side not vertex to side! $\endgroup$
    – user77970
    Oct 31, 2013 at 14:59
  • $\begingroup$ @GregHill: You're right ... I inadvertently (and embarrassingly) included perpendicular bisectors when I wrote that "those 'lines' [are] 'cevians'". Thank you for the correction. I'll note that the perpendicular bisectors of $\triangle ABC$ are the altitudes of $\triangle ABC$'s medial triangle (joining the midpoints of sides) ... so, the lines are cevians of a triangle, just not of the triangle in question. :) $\endgroup$
    – Blue
    Oct 31, 2013 at 15:33
  • 1
    $\begingroup$ @Blue Yes that's true --- and I forgot to qualify my statement by saying that perpendicular bisectors are also cevians when the triangle is isosceles :) $\endgroup$
    – user77970
    Oct 31, 2013 at 20:55

4 Answers 4


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.

  • $\begingroup$ That's... quite obvious, now that I come to think of it! Haha :) Thanks! What about the other lines? $\endgroup$ Oct 18, 2013 at 14:05
  • 3
    $\begingroup$ For the angle bisectors, any point is on an angle bisector if and only if it is equidistant from the two corresponding edges of the triangle. So where two angle bisectors meet is equidistant from all $3$ edges, so is on the third angle bisector. For medians, there are many proofs, all a little longer than a comment. $\endgroup$ Oct 18, 2013 at 14:24
  • $\begingroup$ @AndréNicolas Thanks for that! :) What about the perpendicular bisectors though? And what is the idea behind the median proof? Maybe you don't have to write the entire thing, and I can guess it. :P $\endgroup$ Oct 20, 2013 at 4:34
  • $\begingroup$ Perpendicular bisectors are dealt with in the answer by Daniel Rust. Medians are perhaps easiest to see as meeting at the centroid. $\endgroup$ Oct 20, 2013 at 4:48
  • $\begingroup$ @AndréNicolas So sorry, I meant altitudes! :P $\endgroup$ Oct 20, 2013 at 4:49

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?

Ceva's Theorem

Ceva's Theorem applied to altitudes

enter image description here

  • 1
    $\begingroup$ Thanks for the explanation! I've been a bit caught up, so am unable to read through your answer thoroughly, but I'll get back to it ASAP. PS That's what I was looking for: "Does anyone know a way to explain the idea of ALL triangle concurrences based on something intrinsic about triangles?" $\endgroup$ Nov 10, 2013 at 6:48

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.

Extended Ceva' Theorem

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.)

  • $\begingroup$ Ortha's Theorem ... Most amusing! $\endgroup$
    – user77970
    May 18, 2017 at 15:58
  • $\begingroup$ So Ortha's theorem was Carnot's theorem (you already know it :) ). For me it is kind of a rearrangement (theorem). We take different sets of right triangles so the legs are rearranged. Same rearrangement can be seen in quadrilaterals with perpendicular diagonals. $\endgroup$
    – cosmo5
    Apr 17, 2021 at 5:43

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 $

  • $\begingroup$ I'm not sure I completely understand this answer, but I have not seen this approach to a proof of Ceva's theorem - nice to see another way. $\endgroup$
    – user77970
    Oct 31, 2013 at 15:01
  • $\begingroup$ @GregHill: Take a look at this article by two high school students. This is where I learn all my barycentric coordinates! $\endgroup$ Oct 31, 2013 at 15:03
  • $\begingroup$ Looks like a good article, I'll check it out - thanks! $\endgroup$
    – user77970
    Oct 31, 2013 at 21:03

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .