I recently found out that the result I discovered and proved in my post Cool orthocentre thing is actually a consequence of a known theorem, Brokard’s theorem. I had never heard of this theorem before but apparently it’s well-known to serious geometers! (I guess that means I’m not a serious geometer.) However, the proofs I have found for it online typically employ projective geometry or inversion. These techniques are fine – I just haven’t studied them yet! So at least I managed to find a low-technology proof of Brokard’s theorem, even if it was really long. The pro inversion proofs take like 3 lines – see this IMO maths page (too hardcore for me)

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

*without*having to use Brokard’s Theorem. We construct some new points:

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.

Hello.

Is this DanY?

Where the hell is this place?

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Yeah I’m a Dan Y?

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