Harmonic Divisions - A Powerful & Rarely Used Tool!
by Potla, Feb 8, 2011, 5:56 PM
Long since I have ever posted anything on my blog, and more importantly, I have not posted anything on geometry before (except for a similarity trivia), and therefore this is my first extensive attempt on geometry.
As the title explains clearly, I want to post some useful results and solve some problems related to Harmonic Divisions, because I have not found a very good book or ebook explaining the concepts clearly.
HARMONIC DIVISIONS
Though all of us know what a harmonic division is, if four points
are collinear and if they satisfy
then the four points 
are said to form a harmonic range.
Now let us move onto some theorems.
Theorem 1.
If
is a harmonic range and if 
is the midpoint of 
then 
Proof.
After componendo and dividendo on
we get 
leading to
The converse of this theorem is also obviously true.
![[asy]
dot((0,0));
dot((-5,0));
dot((+5,0));
dot((3,0));
dot((8,0));
draw((-5,0)--(8,0));
label("$A$", (-5,0), S);
label("$O$", (0,0), S);
label("$C$", (3,0), S);
label("$D$", (8,0), S);
label("$B$", (5,0), S);
[/asy]](//latex.artofproblemsolving.com/4/4/2/44296e606a9d3760cba3d9036189d33b10dd1027.png)
Theorem 2.
If
are three concurrent cevians of a triangle 
with 
lying on 
respectively, then 
is a harmonic division.
Proof
Using Menelaus we have,
And using Ceva we obtain
Which jointly lead to
ie the division is harmonic.
![[asy]
size(10cm, 8cm);
label("$A$", (0,5), N);
label("$B$", (-1,0), S);
label("$C$", (5,0), S);
label("$X$", (3,0), S);
label("$Z$", (-0.3, 3.48), NW);
label("$Y$", (2.35,2.65), NE);
label("$T$", (10.82,0), S);
draw((0,5)--(-1,0));
draw((-1,0)--(10.82,0));
draw((5,0)--(0,5));
draw((0,5)--(3,0));
draw((-1,0)--(2.35,2.65));
draw((5,0)--(-0.3,3.48));
draw((-0.3,3.48)--(10.82,0));
[/asy]](//latex.artofproblemsolving.com/e/d/7/ed7573debf7b5e394dd75291202c8bdabe5fe90e.png)
Theorem 3.
If four concurrent lines are such that one transversal cuts them harmonically, then every transversal intersects these four lines to form a harmonic division.
Proof
Assume the lines
pass through and cut by a line 
in respectively. Assume that is harmonic, and therefore let another transversal cut these at 
it is sufficient to check that the four points 
form a harmonic division.
Note that using the sine rule we get
and 
Comparing these two and using the fact that is harmonic, we get 
Now again we can use similar arguments to arrive at
ie 
is harmonic.
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);
draw((-2,0)--(4,0)); draw((2,0)--(10,0)); draw((0,(+12-6*0)/2)--(14.29,(+12-6*14.29)/2)); draw((0,(+24-6*0)/4)--(14.29,(+24-6*14.29)/4)); draw((0,(+60-6*0)/10)--(14.29,(+60-6*14.29)/10)); draw((-2,(+12+6*-2)/2)--(14.29,(+12+6*14.29)/2)); draw((1.33,(+14.29-5.23*1.33)/0.73)--(14.29,(+14.29-5.23*14.29)/0.73));
dot((-2,0),ds); label("$A$", (-2.17,-0.56),S*lsf); dot((4,0),ds); label("$B$", (3.81,-0.47),S*lsf); dot((2,0),ds); label("$C$", (1.67,-0.53),SW*lsf); dot((10,0),ds); label("$D$", (9.83,-0.5),SE*lsf); dot((0,6),ds); label("$O$", (-0.65,6.3),NE*lsf); dot((1.33,9.99),ds); label("$A'$", (1.61,9.93),NE*lsf); dot((2.06,4.76),ds); label("$D'$", (2.2,4.96),NE*lsf); dot((2.39,2.41),ds); label("$B'$", (2.51,2.61),NE*lsf); dot((3.27,-3.8),ds); label("$C'$", (3.53,-3.91),NE*lsf); clip((-5.43,-5.18)--(-5.43,10.64)--(14.29,10.64)--(14.29,-5.18)--cycle);[/asy]](//latex.artofproblemsolving.com/0/8/0/080b9ea965b5a5e9f155182fb068a751ea3ce632.png)
Note: Such four concurrent lines are called to form a harmonic range or a harmonic pencil, denoted as
or sometimes as 
Theorem 4.
Any two of the following statements implore the third statement to be true, too.
Statement 1: The pencil
is harmonic.
Statement 2:
bisects 
internally.
Statement 3: and 
are perpendicular to each other.
Proof.
I will prove that
and the rest is left for the reader to prove.
Refer to the diagram. Draw
such that 
Now,
and 
Using
we get 
ie 
is the midpoint of 
Now, we already have
Therefore bisects 
but since this is perpendicular to 
therefore 
is the external bisector of 
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); pen wwqqff = rgb(0.4,0,1); pen zzzzff = rgb(0.6,0.6,1); pen ffttqq = rgb(1,0.2,0); pen zzwwff = rgb(0.6,0.4,1); pen ffqqww = rgb(1,0,0.4);
draw((-0.82,8.97)--(0.21,8.15)--(1.03,9.18)--(0,10)--cycle,wwqqff); draw((-12,0)--(-5,0),wwqqff); draw((0,10)--(-12,0),wwqqff); draw((0,10)--(-5,0),wwqqff); draw((-5,0)--(12.63,0),wwqqff); draw((0,10)--(12.63,0),wwqqff); draw((0,10)--(-7.92,0),wwqqff); pair parametricplot0_cus(real t){
return (2.79*cos(t)+0,2.79*sin(t)+10);
}
draw(graph(parametricplot0_cus,-2.2406491565943965,-2.0344439357957027)--(0,10)--cycle,ffttqq+linewidth(0.6pt)); pair parametricplot1_cus(real t){
return (2.79*cos(t)+0,2.79*sin(t)+10);
}
draw(graph(parametricplot1_cus,-2.44685437739309,-2.2406491565943965)--(0,10)--cycle,ffttqq); draw((-7.92,0)--(-13.59,-7.16),zzwwff); draw((3.59,7.16)--(-13.59,-7.16),ffqqww);
dot((-12,0),ds); label("$A$", (-12.25,-0.7),NE*lsf); dot((-5,0),ds); label("$B$", (-5.12,-0.75),NE*lsf); dot((0,10),ds); label("$O$", (-0.09,10.39),NE*lsf); dot((-7.92,0),ds); label("$C$", (-8.15,-0.75),NE*lsf); dot((12.63,0),ds); label("$D$", (12.44,-0.84),NE*lsf); dot((-13.59,-7.16),ds); label("$E$", (-13.09,-7.78),NE*lsf); dot((3.59,7.16),ds); label("$F$", (3.78,7.45),NE*lsf); clip((-15,-8.48)--(-15,15.28)--(14.63,15.28)--(14.63,-8.48)--cycle);[/asy]](//latex.artofproblemsolving.com/2/5/2/2527949a8ee00c02b3f356cb50243aba12962fe2.png)
Theorem 5.
If
and 
are harmonic, and if 
concur, then the line 
is also concurrent with these lines.
Proof.
Let
meet 
at some point 
We obtain that 
is harmonic, implying 
is also harmonic. So we get
But also 
so that
so that 
and 
are coincident.
Theorem 6. (NEW)
Let
be a convex quadrilateral. If 
then we must have 
to form a harmonic division.
Proof.
Let
Then from Theorem 2 in 
with cevians 
and transversal 
we observe that the division 
is harmonic.
So the range
is harmonic and using 
as a transversal, we see that 
We are done.
![[asy]
import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwwwff = rgb(0.4,0.4,1); pen ffcccc = rgb(1,0.8,0.8);
draw((0,4)--(-2,0)--(4,0)--(2.02,3.66)--cycle); draw((0,4)--(-2,0),wwwwff); draw((-2,0)--(4,0),wwwwff); draw((4,0)--(2.02,3.66),wwwwff); draw((2.02,3.66)--(0,4),wwwwff); draw((-2,(+8+4*-2)/2)--(6.39,(+8+4*6.39)/2),wwwwff); draw((-2.92,(-14.64+3.66*-2.92)/-1.98)--(4,(-14.64+3.66*4)/-1.98),wwwwff); draw((-2,0)--(2.02,3.66),wwwwff); draw((4,0)--(0,4),wwwwff); draw((0.88,(+4.05-2.9*0.88)/0.26)--(6.39,(+4.05-2.9*6.39)/0.26),wwwwff); draw((-2.92,(-15.29+3.82*-2.92)/-2.95)--(4,(-15.29+3.82*4)/-2.95),wwwwff);
dot((0,4),ds); label("$D$", (-0.23,4.02),W*lsf); dot((-2,0),ds); label("$A$", (-2.17,-0.24),SW*lsf); dot((4,0),ds); label("$B$", (4.02,-0.3),NE*lsf); dot((2.02,3.66),ds); label("$C$", (2.07,3.74),NE*lsf); dot((0.88,5.76),ds); label("$K$", (0.99,5.72),NE*lsf); dot((1.14,2.86),ds); label("$M$", (0.87,2.76),W*lsf); dot((1.05,3.82),ds); label("$Q$", (1.11,3.91),NE*lsf); dot((1.4,0),ds); label("$P$", (1.44,-0.24),SE*lsf); dot((0.36,4.72),ds); label("$R$", (0.3,4.89),N*lsf); clip((-2.92,-1.31)--(-2.92,6.16)--(6.39,6.16)--(6.39,-1.31)--cycle);
[/asy]](//latex.artofproblemsolving.com/f/a/9/fa9e4e30165ba4b64dc3d5eb18e11fad02be7947.png)
Now we have already done some easy theorems which might come in handy here and there, so we need to keep a tab on them. Let's move onto some simple applications.
Note: From now on I might denote as the cross-ratio and 
obviously denotes the case when is harmonic.
Note 2: I assume the reader to have some knowledge on Poles and Polars, Inversion, Homothety and Cross ratio. So hope I will not be too ambiguous while using these notations and notions.
If 
are the altitudes of a triangle 
and if 
meet 
at 
respectively, then show that 
and 
intersect on a point lying on 
Solution
Let meet 
in 
then we have to show that 
are collinear.
Using Theorem 2, we have that
is harmonic, ie 
Similarly we obtain three other relations, and multiplying them out we get
So that using the converse of Menelaus theorem,
is a transversal to the triangle and therefore a straight line.
Note: Even if I used directed segments in this proof, I think it is best to avoid using them.
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen zzttqq = rgb(0.6,0.2,0); pen ffwwww = rgb(1,0.4,0.4); pen qqzzzz = rgb(0,0.6,0.6);
draw((0,8)--(-2,0)--(4,0)--cycle); draw((0,8)--(-2,0),zzttqq); draw((-2,0)--(4,0),zzttqq); draw((4,0)--(0,8),zzttqq); draw((4,0)--(-8,0),wwqqff); draw((-8,0)--(7,-6),wwqqff); draw((0,8)--(7,-6),wwqqff); draw((2.8,2.4)--(-8,0),wwqqff); draw((0,8)--(-2.55,-2.18),wwqqff); draw((0,8)--(0,0),ffwwww); draw((-2,0)--(2.8,2.4),ffwwww); draw((4,0)--(-1.65,1.41),ffwwww); draw((-1.65,1.41)--(7,-6),qqzzzz); draw((2.8,2.4)--(-2.55,-2.18),wwqqff);
dot((-2,0),ds); label("$B$", (-2.56,-0.45),NE*lsf); dot((4,0),ds); label("$C$", (4.44,-0.36),NE*lsf); dot((0,8),ds); label("$A$", (-0.12,8.28),NE*lsf); dot((0,0),ds); label("$D$", (-0.12,-0.52),NE*lsf); dot((2.8,2.4),ds); label("$E$", (2.78,2.66),NE*lsf); dot((-1.65,1.41),ds); label("$F$", (-1.82,1.58),NW*lsf); dot((-2.55,-2.18),ds); label("$F'$", (-2.9,-2.89),NE*lsf); dot((-8,0),ds); label("$D'$", (-8.33,-0.45),S*lsf); dot((7,-6),ds); label("$E'$", (7.22,-6.01),NE*lsf); clip((-9.6,-6.75)--(-9.6,8.99)--(10.03,8.99)--(10.03,-6.75)--cycle);[/asy]](//latex.artofproblemsolving.com/2/7/7/27758bda6b1f5dcb89df92f5c5d5ec80d2dcb2ad.png)
The tangent at a point 
of a circle cuts a diametre at 
; and 
is the perpendicular to 
the line joining to the midpoint of 
cuts at 
Prove that 
Solution
Let us assume
Note that we have, from the similarity of 
and 
that
So using Theorem 4, we obtain that 
is harmonic. So the pencil 
is also harmonic, leading to (suing 
as a transversal) 
is harmonic.
But, this will be harmonic if and only if
But using 
we get 
but 
Therefore 
and
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen ffwwww = rgb(1,0.4,0.4); pen wwwwff = rgb(0.4,0.4,1); pen qqzzzz = rgb(0,0.6,0.6);
draw(circle((0,0),3),ffwwww); draw((-3,0)--(1.82,-3.68),wwwwff); draw((-3,0)--(4.93,0),wwwwff); draw((1.82,2.38)--(1.82,-3.68),qqzzzz); draw((1.82,2.38)--(4.93,0),wwwwff); draw((3.38,1.19)--(1.82,-3.68),wwwwff); draw((1.82,2.38)--(-3,0),wwwwff); draw((3,0)--(1.82,2.38),wwwwff);
dot((0,0),ds); label("$O$", (-0.06,-0.3),NE*lsf); dot((1.82,2.38),ds); label("$P$", (1.88,2.47),NE*lsf); dot((-3,0),ds); label("$A$", (-3.2,-0.25),W*lsf); dot((3,0),ds); label("$B$", (3.03,-0.25),NE*lsf); dot((4.93,0),ds); label("$T$", (4.84,-0.3),NE*lsf); dot((1.82,0),ds); label("$N$", (1.86,-0.28),NE*lsf); dot((3.38,1.19),ds); label("$M$", (3.49,1.18),NE*lsf); dot((1.82,-3.68),ds); label("$Q$", (1.91,-3.76),NE*lsf); clip((-3.91,-4.09)--(-3.91,3.75)--(5.87,3.75)--(5.87,-4.09)--cycle);
[/asy]](//latex.artofproblemsolving.com/1/0/c/10c30de43c867e2172956bc4891c9344dff13aaf.png)
Let 
be a right angled triangle at 
. is a point on 
. Let 
be the midpoint of 
. 
intersects the perpendicular bisector of at 
. Prove that 
.
Solution
Let
We will show that if is a point on extended such that 
Then we must have 
is the perpendicular from onto 
Now let
and 
meet at 
then; since is the midpoint of it is forced that 
is harmonic.
Now this also implies that
is a harmonic range, so that using 
as a transversal to this range it is implored that 
are harmonic.
Now since
we must have 
Therefore using 
we have 
and therefore 
Hence is the midpoint of 
![[asy]
import graph; size(10cm);
draw((0,0)--(4,0));
draw((4,0)--(0,5));
draw((0,0)--(0,5));
draw((0,5)--(2,-2.51));
draw((0,0)--(2,2.5));
draw((0,0)--(2,-2.51));
draw((4,0)--(2,-2.51));
draw((2,0)--(2,-2.51));
label("$A$", (0,0), W);
label("$B$", (4,0), S);
label("$C$", (0,5), N);
label("$D$", (2,2.25), NE);
label("$E$", (2,-2.51), E);
label("$M$", (1,1.25), W);
label("$L$", (2,0), N);
[/asy]](//latex.artofproblemsolving.com/9/8/7/987726bead621b87848e94aaec0f14e99cc53e5a.png)
In acute triangle 
we have points and on sides 
respectively satisfying 
Let the angle bisector of 
meet at 
and are projections of 
and to 
respectively, and 
is the midpoint of 
If the incenter of lies on the circumcircle of 
prove that 
and the incenter of 
are collinear.
Solution(motal)
Let
, 
the incenter of and 
the incenter of . 
so, considering 
with transversal 
, the problem is equivalent to proving that 
. Easy angle chasing leads to 
isosceles. Furthermore, 
concyclic. With similar argument we obtain 
concyclic. 
. Therefore we have that 
is concyclic with diameter 
. In particular we have that both of these facts hold: 
and 
is bisector of 
. Hence considering pencil 
we can conclude that .
We are done.
Sorry I could not attach a figure, I will attach one as soon as possible.
(Desargues) Let and 
be perspective triangles perspective about a point 
If 
and 
then 
are collinear.
Solution
Let us denote
Now we will use a little of cross ratio or projective geometry to prove this fact.
Firstly consider the concurrent points
and 
The corresponding lines 
are concurrent at Therefore using a simple idea of cross-ratio we obtain that
It is also obvious that 
and 
So the points are collinear since lies on the line 
(Why?)
We are done.
Given that is a diameter of a circle and the lines 
are tangents to the circle, prove that 
where 
is the foot of perpendicular from onto and 
Solution
Let
and 
Since
and 
we see that 
therefore bisects 
and also 
So the range 
is harmonic. Using 
as a transversal this leads to the fact that 
are harmonic. So 
is harmonic, and again, using 
as a transversal to this range we have that 
ie is the midpoint of 
These types of midpoint considerations can prove a lot of geometric problems easily, and of course this one is obvious from considering symmedians. Apart from that, harmonic divisions stands out to be a hugely useful and rarely used (in my country, at least) method.
(Proposed in Olympiad Marathon by Fersolve)
Let be the intersection of the diagonals of convex quadrilateral 
The circumcircles
of
and 
meet at and 
Line 
meets the circumcircles of 
and 
at 
and respectively. Prove that is the midpoint of 
(NEW)
![[asy]import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen ffcccc = rgb(1,0.8,0.8); pen fftttt = rgb(1,0.2,0.2);
draw((0,5)--(-2,0)--(6,0)--(2.31,6.49)--cycle); draw((0,5)--(-2,0),qqttff); draw((-2,0)--(6,0),qqttff); draw((6,0)--(2.31,6.49),qqttff); draw((2.31,6.49)--(0,5),qqttff); draw((-2,0)--(2.31,6.49),qqttff); draw((0,5)--(6,0),qqttff); draw(circle((-1.27,2.61),2.71),fftttt); draw(circle((4.51,3.45),3.76),fftttt); draw((-5.22,(-3.97+2.69*-5.22)/-0.39)--(10.02,(-3.97+2.69*10.02)/-0.39),qqttff); draw(circle((1.24,5.62),1.38),fftttt); draw(circle((2,0.44),4.02),fftttt);
dot((0,5),ds); label("$D$", (-0.43,5.18),NW*lsf); dot((-2,0),ds); label("$A$", (-2.34,-0.36),W*lsf); dot((6,0),ds); label("$B$", (6.14,-0.36),NE*lsf); dot((2.31,6.49),ds); label("$C$", (2.4,6.64),NE*lsf); dot((0.85,4.29),ds); label("$O$", (1.18,4.03),NE*lsf); dot((1.24,1.6),ds); label("$M$", (1.37,1.61),NE*lsf); dot((0.49,6.78),ds); label("$T$", (0.56,7.02),NE*lsf); dot((1.99,-3.59),ds); label("$S$", (2.09,-3.45),NE*lsf); clip((-5.22,-4.1)--(-5.22,8.12)--(10.02,8.12)--(10.02,-4.1)--cycle);[/asy]](//latex.artofproblemsolving.com/9/9/e/99e9c6a3f10bfb2c134810ead879301d0a283807.png)
Solution
Let us invert everything about to map points to their primes. By Theorem 6 on the four collinear points 
we have 
Also we have the following relations:
![\[M'T'=\dfrac{MT\times r^2}{OM\cdot OT}, T'O=\dfrac{r^2}{OT}, M'S'=\dfrac{MS\times r^2}{OM\cdot OS}, S'O=\dfrac{r^2}{OS};\]](//latex.artofproblemsolving.com/8/7/0/8704bf25a5f0a05a05e4d04667a87d42cf53a294.png)
And therefore we obtain 
and so, is the midpoint of 
as required.
I will keep updating this post from time to time because as time goes on, the number of problems you come across also goes on.
I will add some exercises about a month later (after my secondary exams end), so, till then; Rejoice!!
References
[1] Cosmin Pohoata; Harmonic Divisions; 2007 Mathematical Reflections
[2] C. Stanley Ogilvy; (1990) Excursions in Geometry, Dover.
[3] Application 1
[4] Application 2
Last Updated On: 7:00 pm UCT; 03 March, 2011.
As the title explains clearly, I want to post some useful results and solve some problems related to Harmonic Divisions, because I have not found a very good book or ebook explaining the concepts clearly.
HARMONIC DIVISIONS
Though all of us know what a harmonic division is, if four points
are collinear and if they satisfy
then the four points 
are said to form a harmonic range.Now let us move onto some theorems.
Theorem 1.
If
is a harmonic range and if 
is the midpoint of 
then 
Proof.
After componendo and dividendo on
we get 
leading to
The converse of this theorem is also obviously true.
![[asy]
dot((0,0));
dot((-5,0));
dot((+5,0));
dot((3,0));
dot((8,0));
draw((-5,0)--(8,0));
label("$A$", (-5,0), S);
label("$O$", (0,0), S);
label("$C$", (3,0), S);
label("$D$", (8,0), S);
label("$B$", (5,0), S);
[/asy]](http://latex.artofproblemsolving.com/4/4/2/44296e606a9d3760cba3d9036189d33b10dd1027.png)
Theorem 2.
If
are three concurrent cevians of a triangle 
with 
lying on 
respectively, then 
is a harmonic division.Proof
Using Menelaus we have,
And using Ceva we obtain
Which jointly lead to
ie the division is harmonic.![[asy]
size(10cm, 8cm);
label("$A$", (0,5), N);
label("$B$", (-1,0), S);
label("$C$", (5,0), S);
label("$X$", (3,0), S);
label("$Z$", (-0.3, 3.48), NW);
label("$Y$", (2.35,2.65), NE);
label("$T$", (10.82,0), S);
draw((0,5)--(-1,0));
draw((-1,0)--(10.82,0));
draw((5,0)--(0,5));
draw((0,5)--(3,0));
draw((-1,0)--(2.35,2.65));
draw((5,0)--(-0.3,3.48));
draw((-0.3,3.48)--(10.82,0));
[/asy]](http://latex.artofproblemsolving.com/e/d/7/ed7573debf7b5e394dd75291202c8bdabe5fe90e.png)
Theorem 3.
If four concurrent lines are such that one transversal cuts them harmonically, then every transversal intersects these four lines to form a harmonic division.
Proof
Assume the lines
pass through and cut by a line 
in respectively. Assume that is harmonic, and therefore let another transversal cut these at 
it is sufficient to check that the four points 
form a harmonic division.Note that using the sine rule we get
and 
Comparing these two and using the fact that
is harmonic, we get 
Now again we can use similar arguments to arrive at
ie 
is harmonic.![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1);
draw((-2,0)--(4,0)); draw((2,0)--(10,0)); draw((0,(+12-6*0)/2)--(14.29,(+12-6*14.29)/2)); draw((0,(+24-6*0)/4)--(14.29,(+24-6*14.29)/4)); draw((0,(+60-6*0)/10)--(14.29,(+60-6*14.29)/10)); draw((-2,(+12+6*-2)/2)--(14.29,(+12+6*14.29)/2)); draw((1.33,(+14.29-5.23*1.33)/0.73)--(14.29,(+14.29-5.23*14.29)/0.73));
dot((-2,0),ds); label("$A$", (-2.17,-0.56),S*lsf); dot((4,0),ds); label("$B$", (3.81,-0.47),S*lsf); dot((2,0),ds); label("$C$", (1.67,-0.53),SW*lsf); dot((10,0),ds); label("$D$", (9.83,-0.5),SE*lsf); dot((0,6),ds); label("$O$", (-0.65,6.3),NE*lsf); dot((1.33,9.99),ds); label("$A'$", (1.61,9.93),NE*lsf); dot((2.06,4.76),ds); label("$D'$", (2.2,4.96),NE*lsf); dot((2.39,2.41),ds); label("$B'$", (2.51,2.61),NE*lsf); dot((3.27,-3.8),ds); label("$C'$", (3.53,-3.91),NE*lsf); clip((-5.43,-5.18)--(-5.43,10.64)--(14.29,10.64)--(14.29,-5.18)--cycle);[/asy]](http://latex.artofproblemsolving.com/0/8/0/080b9ea965b5a5e9f155182fb068a751ea3ce632.png)
Note: Such four concurrent lines are called to form a harmonic range or a harmonic pencil, denoted as
or sometimes as 
Theorem 4.
Any two of the following statements implore the third statement to be true, too.
Statement 1: The pencil
is harmonic.Statement 2:
bisects 
internally.Statement 3:
and 
are perpendicular to each other.Proof.
I will prove that
and the rest is left for the reader to prove.Refer to the diagram. Draw
such that 
Now,
and 
Using
we get 
ie 
is the midpoint of 
Now, we already have
Therefore bisects 
but since this is perpendicular to 
therefore 
is the external bisector of 
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); pen wwqqff = rgb(0.4,0,1); pen zzzzff = rgb(0.6,0.6,1); pen ffttqq = rgb(1,0.2,0); pen zzwwff = rgb(0.6,0.4,1); pen ffqqww = rgb(1,0,0.4);
draw((-0.82,8.97)--(0.21,8.15)--(1.03,9.18)--(0,10)--cycle,wwqqff); draw((-12,0)--(-5,0),wwqqff); draw((0,10)--(-12,0),wwqqff); draw((0,10)--(-5,0),wwqqff); draw((-5,0)--(12.63,0),wwqqff); draw((0,10)--(12.63,0),wwqqff); draw((0,10)--(-7.92,0),wwqqff); pair parametricplot0_cus(real t){
return (2.79*cos(t)+0,2.79*sin(t)+10);
}
draw(graph(parametricplot0_cus,-2.2406491565943965,-2.0344439357957027)--(0,10)--cycle,ffttqq+linewidth(0.6pt)); pair parametricplot1_cus(real t){
return (2.79*cos(t)+0,2.79*sin(t)+10);
}
draw(graph(parametricplot1_cus,-2.44685437739309,-2.2406491565943965)--(0,10)--cycle,ffttqq); draw((-7.92,0)--(-13.59,-7.16),zzwwff); draw((3.59,7.16)--(-13.59,-7.16),ffqqww);
dot((-12,0),ds); label("$A$", (-12.25,-0.7),NE*lsf); dot((-5,0),ds); label("$B$", (-5.12,-0.75),NE*lsf); dot((0,10),ds); label("$O$", (-0.09,10.39),NE*lsf); dot((-7.92,0),ds); label("$C$", (-8.15,-0.75),NE*lsf); dot((12.63,0),ds); label("$D$", (12.44,-0.84),NE*lsf); dot((-13.59,-7.16),ds); label("$E$", (-13.09,-7.78),NE*lsf); dot((3.59,7.16),ds); label("$F$", (3.78,7.45),NE*lsf); clip((-15,-8.48)--(-15,15.28)--(14.63,15.28)--(14.63,-8.48)--cycle);[/asy]](http://latex.artofproblemsolving.com/2/5/2/2527949a8ee00c02b3f356cb50243aba12962fe2.png)
Theorem 5.
If
and 
are harmonic, and if 
concur, then the line 
is also concurrent with these lines.Proof.
Let
meet 
at some point 
We obtain that 
is harmonic, implying 
is also harmonic. So we get
But also 
so that
so that 
and 
are coincident.
Theorem 6. (NEW)
Let
be a convex quadrilateral. If 
then we must have 
to form a harmonic division.Proof.
Let
Then from Theorem 2 in 
with cevians 
and transversal 
we observe that the division 
is harmonic.So the range
is harmonic and using 
as a transversal, we see that 
We are done.![[asy]
import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwwwff = rgb(0.4,0.4,1); pen ffcccc = rgb(1,0.8,0.8);
draw((0,4)--(-2,0)--(4,0)--(2.02,3.66)--cycle); draw((0,4)--(-2,0),wwwwff); draw((-2,0)--(4,0),wwwwff); draw((4,0)--(2.02,3.66),wwwwff); draw((2.02,3.66)--(0,4),wwwwff); draw((-2,(+8+4*-2)/2)--(6.39,(+8+4*6.39)/2),wwwwff); draw((-2.92,(-14.64+3.66*-2.92)/-1.98)--(4,(-14.64+3.66*4)/-1.98),wwwwff); draw((-2,0)--(2.02,3.66),wwwwff); draw((4,0)--(0,4),wwwwff); draw((0.88,(+4.05-2.9*0.88)/0.26)--(6.39,(+4.05-2.9*6.39)/0.26),wwwwff); draw((-2.92,(-15.29+3.82*-2.92)/-2.95)--(4,(-15.29+3.82*4)/-2.95),wwwwff);
dot((0,4),ds); label("$D$", (-0.23,4.02),W*lsf); dot((-2,0),ds); label("$A$", (-2.17,-0.24),SW*lsf); dot((4,0),ds); label("$B$", (4.02,-0.3),NE*lsf); dot((2.02,3.66),ds); label("$C$", (2.07,3.74),NE*lsf); dot((0.88,5.76),ds); label("$K$", (0.99,5.72),NE*lsf); dot((1.14,2.86),ds); label("$M$", (0.87,2.76),W*lsf); dot((1.05,3.82),ds); label("$Q$", (1.11,3.91),NE*lsf); dot((1.4,0),ds); label("$P$", (1.44,-0.24),SE*lsf); dot((0.36,4.72),ds); label("$R$", (0.3,4.89),N*lsf); clip((-2.92,-1.31)--(-2.92,6.16)--(6.39,6.16)--(6.39,-1.31)--cycle);
[/asy]](http://latex.artofproblemsolving.com/f/a/9/fa9e4e30165ba4b64dc3d5eb18e11fad02be7947.png)
Now we have already done some easy theorems which might come in handy here and there, so we need to keep a tab on them. Let's move onto some simple applications.
Note: From now on I might denote
as the cross-ratio and 
obviously denotes the case when is harmonic.Note 2: I assume the reader to have some knowledge on Poles and Polars, Inversion, Homothety and Cross ratio. So hope I will not be too ambiguous while using these notations and notions.
If 
are the altitudes of a triangle 
and if 
meet 
at 
respectively, then show that 
and 
intersect on a point lying on 
Solution
Let
meet 
in 
then we have to show that 
are collinear.Using Theorem 2, we have that
is harmonic, ie 
Similarly we obtain three other relations, and multiplying them out we get
So that using the converse of Menelaus theorem,
is a transversal to the triangle and therefore a straight line.Note: Even if I used directed segments in this proof, I think it is best to avoid using them.
![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.5) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen zzttqq = rgb(0.6,0.2,0); pen ffwwww = rgb(1,0.4,0.4); pen qqzzzz = rgb(0,0.6,0.6);
draw((0,8)--(-2,0)--(4,0)--cycle); draw((0,8)--(-2,0),zzttqq); draw((-2,0)--(4,0),zzttqq); draw((4,0)--(0,8),zzttqq); draw((4,0)--(-8,0),wwqqff); draw((-8,0)--(7,-6),wwqqff); draw((0,8)--(7,-6),wwqqff); draw((2.8,2.4)--(-8,0),wwqqff); draw((0,8)--(-2.55,-2.18),wwqqff); draw((0,8)--(0,0),ffwwww); draw((-2,0)--(2.8,2.4),ffwwww); draw((4,0)--(-1.65,1.41),ffwwww); draw((-1.65,1.41)--(7,-6),qqzzzz); draw((2.8,2.4)--(-2.55,-2.18),wwqqff);
dot((-2,0),ds); label("$B$", (-2.56,-0.45),NE*lsf); dot((4,0),ds); label("$C$", (4.44,-0.36),NE*lsf); dot((0,8),ds); label("$A$", (-0.12,8.28),NE*lsf); dot((0,0),ds); label("$D$", (-0.12,-0.52),NE*lsf); dot((2.8,2.4),ds); label("$E$", (2.78,2.66),NE*lsf); dot((-1.65,1.41),ds); label("$F$", (-1.82,1.58),NW*lsf); dot((-2.55,-2.18),ds); label("$F'$", (-2.9,-2.89),NE*lsf); dot((-8,0),ds); label("$D'$", (-8.33,-0.45),S*lsf); dot((7,-6),ds); label("$E'$", (7.22,-6.01),NE*lsf); clip((-9.6,-6.75)--(-9.6,8.99)--(10.03,8.99)--(10.03,-6.75)--cycle);[/asy]](http://latex.artofproblemsolving.com/2/7/7/27758bda6b1f5dcb89df92f5c5d5ec80d2dcb2ad.png)
The tangent at a point 
of a circle cuts a diametre at 
; and 
is the perpendicular to 
the line joining to the midpoint of 
cuts at 
Prove that 
Solution
Let us assume
Note that we have, from the similarity of 
and 
that
So using Theorem 4, we obtain that 
is harmonic. So the pencil 
is also harmonic, leading to (suing 
as a transversal) 
is harmonic.But, this will be harmonic if and only if
But using 
we get 
but 
Therefore 
and ![[asy]
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen wwqqff = rgb(0.4,0,1); pen ffwwww = rgb(1,0.4,0.4); pen wwwwff = rgb(0.4,0.4,1); pen qqzzzz = rgb(0,0.6,0.6);
draw(circle((0,0),3),ffwwww); draw((-3,0)--(1.82,-3.68),wwwwff); draw((-3,0)--(4.93,0),wwwwff); draw((1.82,2.38)--(1.82,-3.68),qqzzzz); draw((1.82,2.38)--(4.93,0),wwwwff); draw((3.38,1.19)--(1.82,-3.68),wwwwff); draw((1.82,2.38)--(-3,0),wwwwff); draw((3,0)--(1.82,2.38),wwwwff);
dot((0,0),ds); label("$O$", (-0.06,-0.3),NE*lsf); dot((1.82,2.38),ds); label("$P$", (1.88,2.47),NE*lsf); dot((-3,0),ds); label("$A$", (-3.2,-0.25),W*lsf); dot((3,0),ds); label("$B$", (3.03,-0.25),NE*lsf); dot((4.93,0),ds); label("$T$", (4.84,-0.3),NE*lsf); dot((1.82,0),ds); label("$N$", (1.86,-0.28),NE*lsf); dot((3.38,1.19),ds); label("$M$", (3.49,1.18),NE*lsf); dot((1.82,-3.68),ds); label("$Q$", (1.91,-3.76),NE*lsf); clip((-3.91,-4.09)--(-3.91,3.75)--(5.87,3.75)--(5.87,-4.09)--cycle);
[/asy]](http://latex.artofproblemsolving.com/1/0/c/10c30de43c867e2172956bc4891c9344dff13aaf.png)
Let 
be a right angled triangle at 
. is a point on 
. Let 
be the midpoint of 
. 
intersects the perpendicular bisector of at 
. Prove that 
.Solution
Let
We will show that if is a point on extended such that 
Then we must have 
is the perpendicular from onto 
Now let
and 
meet at 
then; since is the midpoint of it is forced that 
is harmonic.Now this also implies that
is a harmonic range, so that using 
as a transversal to this range it is implored that 
are harmonic.Now since
we must have 
Therefore using 
we have 
and therefore 
Hence
is the midpoint of 
![[asy]
import graph; size(10cm);
draw((0,0)--(4,0));
draw((4,0)--(0,5));
draw((0,0)--(0,5));
draw((0,5)--(2,-2.51));
draw((0,0)--(2,2.5));
draw((0,0)--(2,-2.51));
draw((4,0)--(2,-2.51));
draw((2,0)--(2,-2.51));
label("$A$", (0,0), W);
label("$B$", (4,0), S);
label("$C$", (0,5), N);
label("$D$", (2,2.25), NE);
label("$E$", (2,-2.51), E);
label("$M$", (1,1.25), W);
label("$L$", (2,0), N);
[/asy]](http://latex.artofproblemsolving.com/9/8/7/987726bead621b87848e94aaec0f14e99cc53e5a.png)
In acute triangle 
we have points and on sides 
respectively satisfying 
Let the angle bisector of 
meet at 
and are projections of 
and to 
respectively, and 
is the midpoint of 
If the incenter of lies on the circumcircle of 
prove that 
and the incenter of 
are collinear.Solution(motal)
Let
, 
the incenter of and 
the incenter of . 
so, considering 
with transversal 
, the problem is equivalent to proving that 
. Easy angle chasing leads to 
isosceles. Furthermore, 
concyclic. With similar argument we obtain 
concyclic. 
. Therefore we have that 
is concyclic with diameter 
. In particular we have that both of these facts hold: 
and 
is bisector of 
. Hence considering pencil 
we can conclude that .We are done.
Sorry I could not attach a figure, I will attach one as soon as possible.
(Desargues) Let and 
be perspective triangles perspective about a point 
If 
and 
then 
are collinear.Solution
Let us denote
Now we will use a little of cross ratio or projective geometry to prove this fact.Firstly consider the concurrent points
and 
The corresponding lines 
are concurrent at Therefore using a simple idea of cross-ratio we obtain that
It is also obvious that 
and 
So the points are collinear since lies on the line 
(Why?)We are done.
Given that is a diameter of a circle and the lines 
are tangents to the circle, prove that 
where 
is the foot of perpendicular from onto and 
Solution
Let
and 
Since
and 
we see that 
therefore bisects 
and also 
So the range 
is harmonic. Using 
as a transversal this leads to the fact that 
are harmonic. So 
is harmonic, and again, using 
as a transversal to this range we have that 
ie is the midpoint of 
These types of midpoint considerations can prove a lot of geometric problems easily, and of course this one is obvious from considering symmedians. Apart from that, harmonic divisions stands out to be a hugely useful and rarely used (in my country, at least) method.
(Proposed in Olympiad Marathon by Fersolve)Let
be the intersection of the diagonals of convex quadrilateral 
The circumcirclesof
and 
meet at and 
Line 
meets the circumcircles of 
and 
at 
and respectively. Prove that is the midpoint of 
(NEW)![[asy]import graph; size(14cm); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen ffcccc = rgb(1,0.8,0.8); pen fftttt = rgb(1,0.2,0.2);
draw((0,5)--(-2,0)--(6,0)--(2.31,6.49)--cycle); draw((0,5)--(-2,0),qqttff); draw((-2,0)--(6,0),qqttff); draw((6,0)--(2.31,6.49),qqttff); draw((2.31,6.49)--(0,5),qqttff); draw((-2,0)--(2.31,6.49),qqttff); draw((0,5)--(6,0),qqttff); draw(circle((-1.27,2.61),2.71),fftttt); draw(circle((4.51,3.45),3.76),fftttt); draw((-5.22,(-3.97+2.69*-5.22)/-0.39)--(10.02,(-3.97+2.69*10.02)/-0.39),qqttff); draw(circle((1.24,5.62),1.38),fftttt); draw(circle((2,0.44),4.02),fftttt);
dot((0,5),ds); label("$D$", (-0.43,5.18),NW*lsf); dot((-2,0),ds); label("$A$", (-2.34,-0.36),W*lsf); dot((6,0),ds); label("$B$", (6.14,-0.36),NE*lsf); dot((2.31,6.49),ds); label("$C$", (2.4,6.64),NE*lsf); dot((0.85,4.29),ds); label("$O$", (1.18,4.03),NE*lsf); dot((1.24,1.6),ds); label("$M$", (1.37,1.61),NE*lsf); dot((0.49,6.78),ds); label("$T$", (0.56,7.02),NE*lsf); dot((1.99,-3.59),ds); label("$S$", (2.09,-3.45),NE*lsf); clip((-5.22,-4.1)--(-5.22,8.12)--(10.02,8.12)--(10.02,-4.1)--cycle);[/asy]](http://latex.artofproblemsolving.com/9/9/e/99e9c6a3f10bfb2c134810ead879301d0a283807.png)
Solution
Let us invert everything about
to map points to their primes. By Theorem 6 on the four collinear points 
we have 
Also we have the following relations:![\[M'T'=\dfrac{MT\times r^2}{OM\cdot OT}, T'O=\dfrac{r^2}{OT}, M'S'=\dfrac{MS\times r^2}{OM\cdot OS}, S'O=\dfrac{r^2}{OS};\]](http://latex.artofproblemsolving.com/8/7/0/8704bf25a5f0a05a05e4d04667a87d42cf53a294.png)
And therefore we obtain 
and so, is the midpoint of 
as required.I will keep updating this post from time to time because as time goes on, the number of problems you come across also goes on.
I will add some exercises about a month later (after my secondary exams end), so, till then; Rejoice!!
References
[1] Cosmin Pohoata; Harmonic Divisions; 2007 Mathematical Reflections
[2] C. Stanley Ogilvy; (1990) Excursions in Geometry, Dover.
[3] Application 1
[4] Application 2
Last Updated On: 7:00 pm UCT; 03 March, 2011.
This post has been edited 6 times. Last edited by levans, Jan 15, 2016, 3:57 AM
.
since under inversion wrt 
the point at infinity comes to 
is mid-point of 
.
March 2012
December 2011