Extend the altitudes of ABC to meet the circumcircle at P, Q and R.

**PQR is an enlargement of the orthic triangle with a scale factor of 2, and the centre of enlargement is the orthocentre of ABC.**

Proof:

This follows from the fact that the orthocentre is the unique point that is reflected in the sides of a triangle onto the circumcircle.

We will prove this.

Finally, the result does not need modification for obtuse triangles: but again, I am not going to concern myself with the obtuse case, as only minor modification to the proof is necessary.

Let D, E and F be the feet of the perpendiculars dropped from A to BC, B to AC and C to AB respectively.

Let the point where AD intersects the circumcircle (other than A) be P. Let the point where BE intersects the circumcircle (other than B) be Q. Let the point where CF intersects the circumcircle (other than C) be R.

Let the orthocentre of ABC be H.

Angle PCB = angle PAB (same segment)

Angle BDA = 90, so by angles in a triangle, angle ABD = 90 – angle PCB

Angle CFB = 90, so by angles in a triangle, angle BCF = 90 – angle ABD = angle PCB

Angle HDC = angle PDC = 90 and CD is common to triangles DCH and PDC; therefore, DCH and PDC are congruent by ASA congruence.

Thus HD = DP

So H reflects in BC to land at P. We can repeat this for the other vertices.

Therefore PQR is an enlargement of DEF from H, scale factor 2.

It follows that H is also the incentre of PQR. PQR will also share the next property of the orthic triangle (to be posted)

By the way, here is a picture of the obtuse case:

Still works!

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