Triangle Centers


The perpendicular bisectors of a triangle are concurrent . And the point of their concurrence is known as "Circumcenter". Basically, it is represented by 'SS'.

The circumcenter is equidistant from all the three vertices. So, how can we say that the perpendicular bisectors are concurrent and circumcenter is equidistant from all the three vertices. So, we shall prove that. But how??? So, first draw the perpendicular bisectors of sides BCBC and ACAC. Let they intersect at SS. Now, drop a perpendicular from SS onto ABAB. Let this perpendicular meet ABAB at FF. So, it is sufficient to prove that SFSF is the perpendicular bisector of ABAB.

So, observe that BD=DCBD=DC (why?) and BDS=CDS=90o\angle BDS = \angle CDS = 90^{o}. So, ΔBDSΔCDS\Delta BDS \cong \Delta CDS. So, SB=SCSB = SC. Similarly proceeding for ΔAES\Delta AES and ΔCES\Delta CES, we obtain SC=SASC =SA. This implies that SA=SCSA = SC. So, ΔSAC\Delta SAC is isosceles. Then what can we say about the altitude SFSF of ΔSAB\Delta SAB. It is also the perpendicular bisector of the side ABAB.(But why?). So, we have finally proved.\blacksquare

So, let's move onto next topic.


The internal angular bisectors of the angles of the triangle are concurrent. The point of concurrence of these angular bisectors is known as "Incenter". It is represented by the letter 'II'. And the radius of the incircle is the perpndicular distance from Incenter to any one the side.

So what is the special property of this Incenter. Actually, it is equidistant from all the sides of the triangle. So, How can we prove these things as like above? Can we use the similar strategy to that? Of Course, we can prove this using the above strategy.

So,let us first draw the angular bisectors of A\angle A and B\angle B. Let they meet at II. So, if we prove that CICI is the angular bisector of C\angle C, we are done. But how? Let IDID,IEIE and IFIF be the perpendiculars from II onto the sides BCBC, CACA and ABAB.

Observe that, DBI=FBI\angle DBI = \angle FBI and IDB=IFB=90o\angle IDB = \angle IFB = 90^{o}. So, ΔIBFΔIBD\Delta IBF \cong \Delta IBD. So, ID=IFID =IF. Applying similar procedure for ΔIFA\Delta IFA and ΔIEA\Delta IEA, we get IE=IFIE = IF. This implies that ID=IEID = IE. But, IDC=IEC=90o\angle IDC = \angle IEC = 90^{o}. So, ΔIDCΔIEC\Delta IDC \cong \Delta IEC. But what can we infer from this?? Oops, actually we're done.(How? Think yourselves). So, we have finally proved this. Are you feeling bored? I think you won't. So, Let's move on. \blacksquare


So, what is this orthocenter? Can you guess it? Okay, right. Actually, Orthocenter is the point of concurrence of the altitudes of triangle. It is generally represented by the letter 'HH.

So, how to prove this? Apply the same strategy as before. In ΔABC\Delta ABC, let ADAD and BEBE be the altitudes through AA and BB onto BCBC and CACA respectively. Let they meet at the point 'HH'. Extend CHCH to meet ABAB at FF. So, we have to prove that CFCF is perpendicular to ABAB.

Since, AEB=ADB=90o\angle AEB = \angle ADB = 90^{o}, it infers that AEDBAEDB is cyclic quadrilateral. Also, HEC+HDC=90o+90o=180o\angle HEC + \angle HDC = 90^{o} + 90^{o} = 180^{o}. So, HECDHECD is cyclic quadrilateral. Now, BAD=BED=90oCED=90oCHD\angle BAD = \angle BED = 90^{o} - \angle CED = 90^{o} - \angle CHD. But, BAD=90oABC\angle BAD = 90^{o} - \angle ABC. So, ABC=90oBAD=90o(90oCHD)=CHD\angle ABC = 90^{o} - \angle BAD = 90^{o} - (90^{o} - \angle CHD) = \angle CHD . Now, FHD=180oCHD=180oBAC\angle FHD = 180^{o} - \angle CHD = 180^{o} - \angle BAC. So, BFHDBFHD is cyclic. This implies that BFH=180oBDH=90o\angle BFH = 180^{o} - \angle BDH = 90^{o}. Phewww!! We're done. Finally proved it. Did you observe that the orthocenter of ΔABC\Delta ABC is the incenter of ΔDEF\Delta DEF.(But How? Try it yourselves)

Before moving to Centroid, let us first discuss about Ceva's Theorem.

Ceva's Theorem:

Consider three concurrent line ADAD,BEBE and CFCF which are concurrent at PP. Then, BDDC.CEEA.AFFB=1\dfrac{BD}{DC} . \dfrac{CE}{EA} . \dfrac{AF}{FB} = 1

Let us prove this. Draw a line parallel to BCBC through AA. Extend BEBE and CFCF to meet this parallel line at BB' and CC'. We can observe that there are so many similar triangles formed here.

As ΔAFCΔBFC\Delta AFC' \sim \Delta BFC and ΔAEBΔBEC\Delta AEB' \sim \Delta BEC, we have AFFB=ACBC\dfrac{AF}{FB} = \dfrac{AC'}{BC} and CEEA=BCAB\dfrac{CE}{EA} = \dfrac{BC}{AB'}.

As ΔAPCΔCPD\Delta APC' \sim \Delta CPD and ΔAPBΔBPD\Delta APB' \sim \Delta BPD. So, ABBD=PAPD\dfrac{AB'}{BD} = \dfrac{PA}{PD} and ACCD=PAPD\dfrac{AC'}{CD} = \dfrac{PA}{PD}.

This implies that ABBD=ACCD\dfrac{AB'}{BD} = \dfrac{AC'}{CD}. So, BDCD=ABAC\dfrac{BD}{CD} = \dfrac{AB'}{AC'}.

BDDC.CEEA.AFFB=ABAC.BCAB.ACBC=1\dfrac{BD}{DC} . \dfrac{CE}{EA} . \dfrac{AF}{FB} = \dfrac{AB'}{AC'} . \dfrac{BC}{AB'} . \dfrac{AC'}{BC} = 1. Hence, proved.

Let us move onto centroid.


Centroid is the point of concurrence of medians of a triangle. It is generally denoted by 'GG'.

So, it is very easy to prove this. Let ADAD and BEBE be the medians. We have to prove that CFCF is also a median.

From Ceva's Theorem,

BDDC.CEEA.AFFB=1\dfrac{BD}{DC} . \dfrac{CE}{EA} . \dfrac{AF}{FB} = 1

But DD and EE are the midpoints of BCBC and CACA. So, BD=DCBD = DC and CE=EACE = EA.

So, AFFB=1\dfrac{AF}{FB} = 1. It implies that AF=FBAF= FB. So, FF is the midpoint of ABAB which infers that CFCF is the median. So, the three medians of a triangle are concurrent.



Excenter is the point of concurrence of external angular bisector of two vertices and internal angular bisector of the third vertex.

So, Let us prove that these lines are concurrent.

Draw the external angular bisectors of B\angle B and C\angle C. Let they meet at ExE_{x}. We have to prove that AExAE_{x} is the excenter of A\angle A. Draw the perpendiculars ExDE_{x}D, ExE_{x} and ExFE_{x}F from ExE_{x} onto BCBC,CACA and ABAB respectively.

As, FBEx=DBEx\angle FBE_{x} = \angle DBE_{x} and BFEx=BDEx\angle BFE_{x} = \angle BDE_{x}, it infers that ΔBDExΔBFEx\Delta BDE_{x} \cong \Delta BFE_{x}. So, ExD=ExFE_{x}D = E_{x}F. Similarly ExF=ExEE_{x}F=E_{x}E. So, ExD=ExEE_{x}D = E_{x}E. But AFEx=AEEx=90o\angle AFE_{x} = \angle AEE_{x} = 90^{o}. It implies that ΔAFExΔAEEx\Delta AFE_{x} \cong \Delta AEE_{x}. So, BAEx=CAEx\angle BAEx = \angle CAE_{x}. It implies that AExAE_{x} is the internal angular bisector of A\angle A. Hence, proved. \blacksquare

Till now we had studied about Circumcenter, Incenter, Orthocenter, Centroid and Excenter. Let us move onto some what advanced.

Nine Point Circle:

The nine point circle is named so because it passes through the nine points of the circles . These nine points are:

  1. Midpoint of each side.

  2. The of each altitude.

  3. The midpoint of line segment from each vertex of the triangle to the orthocenter.

This means that these nine points are concyclic. But how can we say that they are concyclic? Let us prove it.

Consider ΔABC\Delta ABC. Let ADAD, BEBE and CFCF be the altitudes from AA, BB and CC respectively. Let A,B,C,P,Q,RA',B',C',P,Q,R be the mid points BC,CA,AB,AH,BH,CHBC,CA,AB,AH,BH,CH respectively.

From, ΔACH\Delta ACH, we get BR=12AHB'R = \frac{1}{2}AH and BRAHB'R||AH.

From ΔABH\Delta ABH, we get CQ=12AHC'Q = \frac{1}{2}AH and CQAHC'Q||AH.

So, BR=CQB'R = C'Q and BRCQB'R||C'Q.

But, BCBCB'C' ||BC, QRBC QR||BC and AHBCAH \perp BC.

So, BCQRB'C' || QR and CQAHBCC'Q||AH \perp B'C'. It implies that BCQRB'C'QR is rectangle and so it is cyclic. Similarly, ABPQA'B'PQ and ACPRA'C'PR are cyclic. This forces that A,B,C,P,Q,RA', B', C', P, Q, R are cyclic.

As BEQ=90o\angle B'EQ = 90^{o} and BCQ=90o\angle B'CQ = 90^{o}(Since we proved before.) It implies that BEQ=BCQ\angle B'EQ = \angle B'CQ. It forces that EE is also concyclic with these points. Similarly, D,FD,F are also concyclic with these points. So, A,B,C,A,B,C,P,Q,RA,B,C,A',B',C',P,Q,R are concyclic. Hoofff!!! Done. \blacksquare

Till now we have discussed about Circumcenter, Incenter, Orthocenter, Centroid, Excenter, Nine point center. Now, let us discuss about

  1. Radius of Nine Point Circle.

  2. Relation between Circumcenter, Nine Point Center, Orthocenter and Centroid.

Radius of Nine Point Circle:

From the fact that radius of circumcircle of rectangle is half the diameter of the rectangle. It is sufficient to find the length of APA'P. Because APA'P is the diagonal of the rectangle ABPQA'B'PQ which are on the nine point circle.

Let BC=aBC = a, AC=bAC=b and AB=cAB =c. And let A,B,CA,B,C represent BAC,ABC,ACB\angle BAC, \angle ABC , \angle ACB. Observe that CDHECDHE is cyclic. So, AHE=180oDHE=C\angle AHE = 180^{o} - \angle DHE = C.

AH=AESinAHE=ABCosBAESinC=cSinCCosA=2RCosAAH = \dfrac{AE}{Sin \angle AHE} = \dfrac{AB Cos \angle BAE}{Sin C} = \dfrac{c}{Sin C} Cos A = 2RCos A

But, since angle subtended by the arc at the center is double the angle subtended by the arc at any point on the circle, BSC=2A\angle BSC = 2A. As BSA=CSA\angle BSA' = \angle CSA'. It implies that CSA=A\angle CSA' = A. So, SA=SACosA=RCosA=12AH=APSA' = SA CosA = R CosA = \frac{1}{2}AH = AP.(where 'R' is the circumradius.)

Now, ADBCAD \perp BC and SABCSA' \perp BC, it forces that ADSAAD||SA'. And, as we proved before that AP=SAAP = SA', it infers that APASAPA'S is a parallelogram. So, AP=SA=RAP = SA = R. So the radius of the nine point circle is equal to half the length of APAP which is equal to 12R\frac{1}{2} R. Finally, achieved it. \blacksquare

Relation between 'H', 'N','G' and 'S' :

First let us prove that centroid divides the median in the ratio of 2:12:1.

Consider ΔABC\Delta ABC. Let AD,BE,CFAD, BE, CF be the medians and let GG be the centroid. Reflect A,G,EA,G, Ew.r.t the line BCBC and let them be A,G,HA',G',H. As BF=CHBF = CH and BFCHBF||CH. It implies that BFCHBFCH is parallelogram. So, FCBHFC||BH. But since AF=FBAF = FB. It infers that AG=GGAG = GG'. But GG=12GDGG' = \frac{1}{2}GD. So, AG=12GDAG = \frac{1}{2}GD and it infers our result that AG:GD=2:1AG:GD = 2:1.

So, how far does this result help us?

Before starting this, Did you observe that nine point center is actually the circumcenter of the triangle formed by joining the midpoints of the sides of the triangle?(But how? Think yourselves.). Let us move on.

Actually, in any triangle the circumcenter, orthocenter, centroid and nine point center lie on the same line. And this line joining them is known as Euler's line.

So, the proof for this very interesting. First, observe that BCBCBC||B'C' and ASBCA'S \perp BC and it implies that BCASB'C' \perp A'S. Similarly ABCSA'B' \perp C'S. It implies that SS is the orthocenter of ΔABC\Delta A'B'C' which is the circumcenter of ΔABC\Delta ABC.

Now, since AG=2GDAG = 2GD. Let us do operation. By squeezing AGAG to half and reflecting it w.r.t GG, we get AA'. Similarly, doing this operation to BB and CC, we get BB' and CC'. So, circumcenter of ΔABC\Delta ABC transforms to circumcenter of ΔABC\Delta A'B'C', i.e. SS transforms to NN. So, 12SG=NG\frac{1}{2}SG = NG. We have proved that SS, GG and NN are collinear.

Next, Do the same operations. So, Orthocenter of ΔABC\Delta ABC transforms to orthocenter of ΔABC\Delta A'B'C',i.e. HH transforms to SS. It infers that 12HG=SG\frac{1}{2}HG = SG. \blacksquare

So, from the above two facts, it implies that HN:NG:GS=3:1:2HN:NG:GS = 3:1:2. Pheww!! Finally the note is completed.

In this note I had proved all the proofs only considering that the triangle is acute angled. So, it's your turn to try them to prove for obtuse angled triangle.

So, that's all for now. I will post some more advanced topics in a few days or more.

Note by Surya Prakash
5 years, 10 months ago

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Which will be the next topic?

Shivam Jadhav - 5 years, 10 months ago

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May be next topics will be:

  1. Stewart's Theorem 2.Alternate Segment theorem 3.Pedal line theorem 4.Ptolemy's Theorem and some more.

Surya Prakash - 5 years, 10 months ago

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Thanks! There is a lot of good information here! Can you add them to the respective wikis of Triangles - Incenter, Triangles - Circumcenter, Ceva's Theorem , Triangles - Centroid, Triangles - Orthocenter , Nine Point Circle etc?

Calvin Lin Staff - 5 years, 10 months ago

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Of Course!! I will add them Sir.

Surya Prakash - 5 years, 10 months ago

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Hats off! Excellent note! :)

Nihar Mahajan - 5 years, 10 months ago

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Thank you

Surya Prakash - 5 years, 10 months ago

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@Surya Prakash Thanks man! This really helped a lot! :D

Mehul Arora - 5 years, 9 months ago

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this is really helpful. Have you posted other notes too?

Dev Sharma - 5 years, 9 months ago

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great service for whom these notes are unknown!

Prasath M - 5 years, 4 months ago

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