Triangle 9 points on circle

In any triangle ABC,

the mid-points D,E,F of the sides, the feet L,M,N of the altitudes, and the points P,Q,R halfway between the vertices and orthocentre Hall nine lie upon a unique circumference whose centre G is the midpoint of the line segment joining orthocentre H to circumcentre O of ABC and the radius of which is half the radius of the circle that goes through the vertices of ABC.

Let ABC be any plane triangle

Let L,M,N be the feet of the altitude drawn from A,B,C

Let D,E,F be the mid-point of the sides BC, AC,AB

Perpendiculars to the side at D,E,F meet at O, which is equidistant from A,B and C

The lines of the altitudes meet at one point, H, known as the orthocentre.

Let P,Q,R be the points on the altitudes which are half way between orthocentre H and the vertices

Altitudes of triangle

Points P,L,D are not generally collinear

So construct a circle through these points (call the circle K).

<PLD = 1 rt. angle (Construct)

So

PD is diameter of circle K. (because <PLD is a right angle)

PE is parallel to HC (P,E are mid-points of AH,AC of tri.AHC)

Similarly ED is parallel to AB (E,D are mid-points of AC,BC of tri. ABC)

NC is perpendicular to AB (NC is the altitude to side AB from C)

HC is extension of NC (Construct to form orthocentre - see above)

HC is perpendicular to AB (Construct NC being altitude)

So PE is perpendicular to AB (PE being parallel to HC)

So PE is perpendicular to ED (ED being parallel to AB)

So <PED is a right angle

So E lies on the circle K

(Already shown L,D lie on circle and PD is diameter and angle subtended by diagonal is a right angle)

Similarly it can be shown that

FD is parallel to AC,

FP is parallel to BM (in tri. ABH P,F is mid-point of AH,AB)

FP is perpendicular to FD (BM is altitude of B upon AC extended)

So F E lies on the circle K (using same logic as for E above)

We have now shown that P,L,D,E,F lie on the single circle K.

Still need to show that points M,N,Q,R lie on same circle.

If we had started with the circle through points QME in place of PLD

So construct a circle through points QME (call the circle J)

Then

QE is diameter (equiv to PD, <QME rt. ang. equiv. to <PLD)

QD is parallel to HC (Q,D mid-points in tri. HBC)

PE is parallel to HC (P,E mid-points in tri. HAC)

QP is parallel to BA (Q,P mid-points in tri. HAB)

HC is perpendicular to BA (NC being part of HC, NC is alt.)

PE is perpendicular to BA (shown prev. PE is parallel to HC)

So . PE is perpendicular to QP

So . P must lie on circle J (<QPE=1 rt. ang. and QE is diam,)

Similarly

DE is parallel to BA (D,E mid-points in tri. ABC)

HC is perpendicular to BA (NC being part of HC, NC is alt.)

So . QD is perpendicular to BA (shown prev. QD is parallel to HC)

So . QD is perpendicular to DE (shown prev. DE is parallel to BA)

So . D must lie on circle J (<QDE=1 rt. ang. and QE is diam,)

So . P,M,Q,D,E all lie on the single circle J

But it has been shown prev. that P,L,D,E,F lie on the single circle K

and as P,D,E are common to both circles

circle J is the same as circle K (only one circle can pass through 3 points)

We have now shown that P,L,D,E,F,M,Q lie on the single circle K

Still need to show that points R,N lie on circle K.

If we had started with the circle through points FNR in place of PLD or QME

So construct a circle through points FNR (call the circle V)

Then

FR is diameter (equiv to PD, QE <FNR rt. ang. equiv. to <PLD)

FE is parallel to BC (F,E mid-points in tri. ABC)

RE is parallel to AH (R,E mid-points in tri. CAH)

RE is parallel to AL (AH is extension of AL)

AL is perpendicular to BC (AL is altitude of A in tri. ABC))

AL is perpendicular to FE (shown prev. FE is parallel to BC)

So . RE is perpendicular to FE (shown prev. RE is parallel to AL)

So. R must lie on circle V (<REF=1 rt. ang. and FR is diam,)

Similarly

RD is parallel to BH (R,D mid-points in tri. BHC)

FD is parallel to AC (F,D mid-points in tri. ABC)

MC is perpendicular to BH (MC is altitude of B in tri. ABC)

So. FD is perpendicular to BH (shown prev. FD is parallel to MC)

So. FD is perpendicular to RD (shown prev. RD is parallel to BH)

So. D must lie on circle V (<FDR=1 rt. ang. and FR is diam,)

So F,N,R,E,D all lie on the single circle V

But it has been shown that P,L,D,E,F,M,Q lie on the single circle K

and as D.E.F are common to both circles

circle V is the same as circle K (only one circle can pass through 3 points)

We have now shown that all nine points P,M,Q,F,L,D,E,R,N lie on the single circle K

Created 11th August 2008