In a triangle ABC, the orthic triangle is formed by drawing in the altitudes and connecting the points where each altitude meets the opposite side.

It turns out that the orthic triangle has several cool properties.

The first of these properties is:

**The orthocentre of ABC is the incentre of the orthic triangle.**

This property is quite well-known. It holds for all acute triangles.

If ABC is obtuse, the result it very similar (see bottom): I will prove the result for acute triangles here. If ABC is right-angled the orthic triangle is degenerate and all my posts on the orthic triangle become moot.

The acute case:

Let D be the foot of the perpendicular from A to BC; let E be the foot of the perpendicular from B to AC; and let F be the foot of the perpendicular from C to AB. Let H be the orthocentre of ABC.

We seek to show that the altitudes of ABC bisect the angles of DEF.

Firstly since angle BFH = angle BDH = 90, these angles are supplementary; therefore, BDHF is cyclic by the converse of opposite angles in a cyclic quadrilateral.

The same applies to quadrilateral DCEH.

Also, angle BFC = angle BEC = 90; therefore, by the converse of angles in the same segment, BCEF is cyclic.

So:

Therefore:

Angle FDH = angle FBH (same segment in BDHF)

Angle FBH = angle HCE (same segment in BCEF)

Angle HCE = angle HDE (same segment in DCEH)

Hence angle FDH = angle HDE

So AD bisects angle FDE.

We can repeat this twice more to show that all three altitudes bisect the angles of the orthic triangle.

Therefore H is the incentre of DEF.

By the way, here is the obtuse case:

As you can see, A and H (the orthocentre) have simply exchanged places. The result is otherwise the same and can be proven in the same way.

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