Anyway, let’s see Brokard’s theorem in action! Here’s something I noticed:

**Let ABC be an acute-angled triangle with AB < AC. D and E are the feet of the altitudes from B and C respectively, and H is the orthocentre of triangle ABC. DE and CB produced meet at P and the midpoint of DE is M. Then AMHP is a cyclic quadrilateral.**

**Proof**

Let N be the midpoint of BC. Then observe that since ∠BED = ∠BDC = 90, E and D lie on the circle with diameter BC by the converse of Thales’ theorem. Thus BEDC is a cyclic quadrilateral with centre N.

Now we can apply Brokard’s theorem, which tells us that N is the orthocentre of triangle AHP – or equivalently, that H is the orthocentre of triangle APN. Therefore ∠HNP = ∠HAP.

Finally, observe that triangles HDE and HCB are similar (their angles correspond under the same segment theorem). Thus triangles HME and HNB are similar as well, implying that ∠HNB = ∠HME.

So ∠HMP = ∠HME = ∠HNB = ∠HNP = ∠HAP.

As ∠HMP = ∠HAP, AMHP is cyclic by the converse of the same segment theorem.

**EDIT
**Here’s a way to show that H is the orthocentre of triangle APN

Here, K is the foot of the altitude from A, and Q is the foot of the perpendicular dropped from H to AN.

First note that ∠AEH + ∠AQH = 90 + 90 = 180 so AEHQ is cyclic by the converse of ‘opposite angles in a cyclic quad are supplementary’. Similarly AHQD is cyclic by the converse of the same segment theorem since ∠ADH = ∠AQH = 90. Thus A, E, H, Q and D are concyclic, so in particular, DEHQ is a cyclic quad.

KHQN is also cyclic by the converse of ‘opposite angles in a cyclic quad are supplementary’ as ∠HKN + ∠HQN = 90 + 90 = 180.

Finally, EKND is cyclic as E, K, N and D all belong to the nine-point circle of triangle ABC.

Now by the radical axis theorem, the radical axes of DEHQ, KHQN and EKND must be concurrent; ie, DE, QH and NK concur. So P, H and Q are collinear.

Thus PH is perpendicular to AN; since AH is perpendicular to PN, H is the orthocentre of triangle APN.

**ABC is a triangle with AB = 2BC. M is the midpoint of AB and P is on the segment AC such that AC = 3AP. Then PMC = 90.
**

**Proof 1
**Choose D on AB produced beyond B such that MB = BD.

Then as BM = BC = BD, B is the circumcentre of triangle MCD. Since B lies on MD, MD is a diameter of MCD’s circumcircle so ∠MCD = 90.

But CD is parallel to PM by the converse of the intercept theorem as AC = 3AP and AD = 3AM.

Hence ∠PMC = 90 by alternate angles.

Choose D on BC produced beyond C such that BC = CD. Let Q be the midpoint of PC, so AP = PQ = QC.

Then by the converse of the intercept theorem, BQ is parallel to MP and MC is parallel to AD.

But now consider triangle ABD. AC is a median and Q is the point on the median such that AQ = 2QC. Thus Q is the centroid of triangle ABD.

Also, BD = 2BC = AB so triangle ABD is isosceles. Thus the median from B is perpendicular to AD, so BQ is perpendicular to AD.

Using our parallelisms, this implies MP and MC are perpendicular as well.

Let the reflection of H across BC be H’. It is well known that H’ lies on the circumcircle of triangle ABC.

Then OP + PH = OP + PH’. The length of the path OPH’ is at least the straight-line distance from O to H’, which is R, with equality if and only if O, P and H’ are collinear (ie if ∠OPB = ∠HPC). ]]>

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I will just handle the case where E is on the opposite side of AD to C. You would technically have to consider the cases when E is between AD and C, B and C, and to the left of B, but they are almost identical proofs.

This is actually pretty trivial – I just thought it’s quite nice. ∠DBE = ∠DAF (same segment) – this is in fact enough to guarantee the circles are tangent to each other. To see this, choose G on BD extended such that FG is parallel to BE:

Then triangle DFG is obtained from triangle DEB by an enlargement of negative scale factor, centre D – their circumcircles are thus related in the same way and so they are tangential at D. Since ∠DBE = ∠DAF and ∠DBE = ∠DGF (alternate angles), ∠DAF = ∠DGF so A lies on the circumcircle of triangle DGF (converse of same segment) whence the result is established.

Alternatively, draw a line tangent to the circle DAF at D.

It is straightforward to show that this line is also tangent to circle BDE at D using the alternate segment theorem and its converse, from which the result also follows. ]]>

Let the circumcircles of triangles BCT and DAT meet at X.

∠TXC = ∠TBC (same segment)

∠DBC = ∠DAC (same segment)

∠DAT = ∠DXT (same segment)

So TX is the internal angle bisector of ∠DXC. (The red angles above show what is going on)

Now produce CX to intersect the circle ABCD again at Y.

We have ∠DXC = 2∠TXC

But ∠TXC = ∠TBC (same segment) as before

And ∠DBC = ∠DYX (same segment)

So ∠DXC = 2∠DYX

Since ∠DXC is an exterior angle of triangle DYX, ∠DXC = ∠DYX + ∠XDY = 2∠DYX so ∠XDY = ∠DYX

Hence triangle DYX is isosceles, with YX = DX. So the perpendicular bisector of DY passes through X. But the perpendicular bisector of DY also passes through O as DY is a chord of the circle ABCD. So OX is the perpendicular bisector of DY and since triangle DYX is isosceles, it follows that OX bisects ∠YXD. Hence OX is the external angle bisector of ∠DXC, so OX and XT are perpendicular.

Also note that ∠DOC = 2∠DYX (angle at centre is twice angle at circumference) so ∠DOC = ∠DXC.

Hence DOXC is a cyclic quadrilateral by the converse of the same segment theorem.

Now let us extend TX to a point S beyond X, just to aid with explanation.

Firstly, ∠BXS = ∠BCA (exterior angle of cyclic quad)

Similarly ∠SXA = ∠BDA (exterior angle of cyclic quad)

Since ∠BCA = ∠BDA (same seg), we can write ∠BXA = ∠BXS + ∠SXA = 2∠BCA

But ∠BOA = 2∠BCA (angle at centre is twice angle at circumference)

Thus ∠BOA = ∠BXA so BXOA is cyclic by the converse of the same segment theorem.

Now PC × PD = PB × PA by power of a point. Thus P has equal power with respect to the two pink circles, so it must lie on their radical axis. Hence P lies on OX.

Also, QC × QB = QD × QA (by power of a point), so it has equal power with respect to the two new pink circles. Thus Q lies on their radical axis, XT.

Thus we have shown that OXP and XTQ are straight lines, and that ∠OXT = 90. Thus QT is perpendicular to the line OP.

But we can repeat exactly the same steps, but considering the circumcircles of triangles BTA and DTC, to show that BT is perpendicular to the line OQ.

Thus, since T lies at the intersection of two altitudes, it is the orthocentre of triangle OPQ.

Since BP is parallel to QC, ∠PQC = ∠MPB.

So ∠PQC + ∠CAP = 180. Hence PQCA is a cyclic quadrilateral by the converse of ‘opposite angles in a cyclic quad add up to 180’

Since M is the midpoint of PQ and N is the midpoint of PA, triangle PMN is taken to triangle PQA by an enlargement, centre P and scale factor 2; therefore, ∠QAP = ∠MNP.

Also, ∠CPQ = ∠CAQ (same segment).

∠CAQ + ∠QAP = ∠CAP so ∠CPQ + ∠MNP = ∠CAP.

So 180 = ∠MPB + ∠CAP = ∠MPB + ∠CPQ + ∠MNP = ∠CPB + ∠MNP. ]]>

Then BP = B’P, and AP is the perpendicular bisector of BB’ so triangle ABB’ is isosceles with AB = AB’.

AB + BP = PC = B’P + B’C = BP + B’C

AB = B’C so AB’ = B’C, which makes triangle AB’C isosceles.

∠ABC = ∠AB’B as triangle ABB’ is isosceles.

∠AB’B = ∠BCA + ∠B’AC (by exterior angle of triangle)

But ∠BCA = ∠B’AC as AB’C is isosceles so ∠AB’B = 2∠BCA

So ∠ABC = 2∠BCA ]]>

Let triangle ABC satisfy BC = 3, CA = 4 and AB = 5.

**1. Radius of incircle r = 1**

Let the incircle touch sides BC, CA and AB at D, E and F respectively. Let the incentre be I.

BD + DC + CE + EA + AF + FB = 3 + 4 + 5 = 12

But BD = FB, DC = CE, EA = AF (equal tangents)

So 2(FB + DC + AF) = 12

FB + DC + AF = 6

DC = 6 – (AF + FB) = 6 – 5 = 1 = CE

∠DCE = 90, and ∠IDC = ∠IEC = 90 (radius is perpendicular to tangent), so ∠DIE = 90 by angles in a quadrilateral. Also DC = CE so CDIE is a square. So r = ID = CE = 1.

Alternatively, if you know the formula Area = rs, then this gives r immediately, because area and semiperimeter are both 6. However, the lengths of the tangents will become useful to us later. It is easy to show now that BD = FB = 2 and EA = AF = 3.

(Almost as interestingly, the exradii are 2, 3, and 6.)

**2. Let O be the circumcentre of ABC. Then ∠AIO = 45 and ∠BIO = 90
**Since the triangle is right-angled, O is the midpoint of AB. So the circumradius R is 5/2.

FB = 2 so OF = 1/2. FI = 1 so OI = √5/2 by Pythagoras. (If you know it you could use OI² = R(R – 2r) to get this)

Also, since BF = 2 and FI = 1, by Pythagoras, BI = √5

But since BI² + OI² = 5 + 5/4 = 25/4 = (5/2)² = OB², ∠BIO = 90 by the converse of Pythagoras.

(Alternatively, OI² = OF × OB which means OI is tangent to the circumcircle of BIF by the converse of tangent-secant theorem. But because ∠BFI = 90, BI is diameter of BIF’s circumcircle by the converse of Thales, so ∠BIO is 90 because tangent is perpendicular to diameter)

Also ∠ABC + ∠BAC = 90, so ∠ABI + ∠BAI = 45. So by angles in a triangle, ∠AIB = 135 so ∠AIO = 45.

There is a ‘converse’;

We first establish that A and B are on opposite sides of IO.

If they were on the same side of OI, then ∠AIB would be 45. Then ∠IBA + ∠IAB = 135 so ∠ABC + ∠BAC = 270, which clearly violates ‘angles in a triangle sum to 180’. (In general it is true that the angle at the incentre is always strictly greater than 90)

So they are on opposite sides of OI. Now the proof is essentially as before but in reverse: ∠AIB = 135 so ∠ACB = 90, so O is the midpoint of AB (using the converse of Thales)

Let the inradius be 1 unit.

As before, OI² = OF × OB, because ∠OIB = ∠BFI = 90, so R² – 2Rr = OF × R.

So OF = R – 2r = R – 2

BF = 2

Also BI² = R² – OI² = 2Rr = 2R (from triangle BIO)

r² + BF² = 1 + 4 = BI² (from triangle BFI)

5 = 2R

R = 5/2

AB = 5

BD = BF = 2; AE = AF = 3

Also because ∠BCA = 90, DIEC is a square again so EC = DC = r = 1 which gives BC = 3, CA = 4 (in units)

**3. Let DI meet EF at G. Then G is the centroid of ABC.**

DG is perpendicular to BC.

For any triangle, AG would pass through the midpoint of BC; this is fairly well-known. G is not always the centroid, however. This fact has no converse – the 3-4-5 triangle is not the only triangle with this property.

Let DG cut AB at X and let BC and FE meet at Y. Let Z be the foot of the perpendicular from F to BC.

Clearly BCA is taken to BDX by an enlargement, centre B, scale factor 2/3 (as BD = 2 and BC = 3), so DX = 8/3.

Also BCA is taken to BZF by an enlargement, centre B, scale factor 2/5 (as BF = 2 and BA = 5), so FZ = 8/5. Also BZ = 6/5 so DZ = 3 – 1 – 6/5 = 4/5

Now let us compute DG.

As FZ = 8/5 and CE = 1, YCE is taken to YZF by an enlargement, centre Y, scale factor 8/5.

So FZ/ZY = EC/CY.

8CY = 5CY + 9

CY=3

So the enlargement taking ECY to GDY, centre Y, has scale factor DY/CY = 4/3. So GD = 4/3 × EC = 4/3

DG is half of DX so G is the midpoint of DX.

Now an enlargement, centre B, scale factor BC/BD = 3/2, takes BDX to BCA. So it takes the midpoint of DX to the midpoint of CA. Thus BT is a median of ABC. Also BG : GT = 2 : 1 so G is the centroid as desired.

**4. ABC is the only triangle with integer side lengths for which its perimeter is twice its area.
**We know that the area A of a triangle with side lengths a, b and c is given by Heron’s formula:

where s is the semiperimeter

Also, drawing the incircle of the triangle,

we see that if the lengths a, b and c form a triangle, we can substitute a = x + y, b = y + z, c = z + x for some positive real numbers x, y, and z.

This gives us nicer equations:

So if 2A = P, then

Now

So x + y + z is either a positive integer, or 1/2 more than an integer.

If it is 1/2 more than an integer, then, by subtracting integers x + y, y + z, and z + x in turn, we learn that x, y and z are all 1/2 more than an integer.

But then xyz is 1/8 more than an integer, whereas x + y + z is 1/2 more than an integer. This is impossible so x + y + z is an integer. By subtracting integers x + y, y + z, and z + x in turn, we learn that x, y and z are all integers. So x, y and z are all positive integers.

Clearly, xyz becomes much larger than x + y + z as x, y and z increase, so this suggests a size argument to bound the number of cases to check.

x and y + z are positive so yz – 1 is positive. Hence we can multiply by yz – 1:

Assume WLOG that

Now if y = 1, this bound doesn’t do anything for us, but

z – 1 divides into z + 1 which gives z = 2, x = 3 or z = 3, x = 2. (If z > 3 then z – 1 does not divide z + 1.)

Otherwise, we can use the bounding. If y = 2, we need to check z = 2 (does not give an integer x) and z = 3 (gives x = 1).

If y > 2 then there are no more solutions as

So our solutions are (x, y, z) = (3, 1, 2), (2, 1, 3), (1, 2, 3), all of which yield {a, b, c} = {3, 4, 5}.

This does indeed give 2A = P.

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Let BO meet circle ABC again at T.

Since BT is a diameter of the circle ABC, ∠BAT = 90 and ∠BCT = 90 by Thales.

CH is perpendicular to AB since H is the orthocentre; and as TA is also perpendicular to AB, TA and CH are parallel.

Similarly, AH and TC are both perpendicular to BC and hence they are parallel.

Thus TCHA is a parallelogram.

So AH = TC.

Also, as BC = 2BM and BT = 2BO, triangle BCT is an enlargement of triangle BMO, centre B and scale factor 2. Hence TC = 2OM so AH = 2OM.

**Proof 2.
**The next three proofs make use of results concerning the nine-point circle and the Euler line.

Recall that the nine-point circle of a triangle passes through: the midpoint of each side (shown orange below); the feet of the three altitudes (green); and the midpoints of the line segments joining each vertex to the orthocentre (red). The orthocentre itself is shown as grey.

Consider enlarging nine-point circle, with the orthocentre as the centre of enlargement, and scale factor two. The red points move away from the orthocentre to land at the vertices of the triangle. Hence this enlargement enlarges the nine-point circle to the triangle’s circumcircle. Therefore, the orthocentre reflects across any point on the nine-point circle to land on the circumcircle.

Returning to our original diagram, let D be the foot of the altitude from A. Denote the reflections of H across D and M as X and Y respectively. As established, X and Y lie on triangle ABC’s circumcircle, because D and M lie on the nine-point circle.

Since X and Y are reflections of H across D and M respectively, triangle HXY is an enlargement of triangle HDM, scale factor 2 and centre H. As AD is an altitude, ∠HDM = 90, so ∠HXY = 90.

Hence, by the converse of Thales, AY is a diameter of triangle ABC’s circumcircle, so O is the midpoint of AY.

Therefore, since HY = 2MY and AY = 2OY, triangle OMY enlarges to triangle AHY, centre Y and scale factor 2. Therefore 2OM = AH.

Let K be the midpoint of AH and let D be the foot of the altitude from A as before. The nine-point circle passes through K, D and M.

Since AD is an altitude, ∠KDM = 90 so KM is a diameter of the nine-point circle by the converse of Thales. Thus N is the midpoint of KM.

Since N bisects OH and KM, OMHK is a parallelogram. Therefore OM = KH so 2OM = AH.

**Proof 4.
**This final proof is the shortest and makes use of the Euler line. That is, the centroid, G, lies on OH such that OG : GH = 1 : 2.

Since G lies on AM and OH, G is the point where AM and OH meet.

Since OG : GH = 1 : 2 and MG : GA = 1 : 2 (since the centroid divides each median in this ratio), the triangle OGM enlarges to triangle HGA, scale factor -2, centre G. OM is taken to AH by this enlargement so 2OM = AH.