# Two Perspective Triangles

Given ΔABC, the medial triangle M_{a}M_{b}M_{c} joins the midpoints of its sides, while the orthic triangle H_{a}H_{b}H_{c} joins the feet of its altitudes. Triangles ABC and M_{a}M_{b}M_{c} are homothetic in their common centroid, such that all of their remarkable points, lines and triangles are also homothetic. As a weak consequence, the corresponding objects of a triangle and its medial triangle are perspective. In particular, this is true for their orthic triangles.

Vladimir Nikolin observed (June, 2012) a simple, yet a non-trivial fact:

The medial triangles of a triangle and of its orthic triangle are perspective.

In other words, lines M_{a}N_{a}, M_{b}N_{b}, M_{c}N_{c}, where N_{a}, N_{b}, and N_{c} are the midpoints of the orthic triangle H_{a}H_{b}H_{c}, are concurrent. Furthermore, they are concurrent in the 9-point center of ΔABC.

### Proof

As the feet of two altitudes, H_{a} and H_{b} lie on the semicircle centered at M_{c}, implying that _{a}M_{c} = H_{b}M_{c}_{a}M_{c}H_{b} isosceles so that the perpendicular bisector of the base H_{a}H_{b} passes through the vertex M_{c}. Since H_{a}H_{b} is a chord in the 9-point circle of ΔABC, the 9-point center N of ΔABC lies on M_{c}N_{c}. For the same reason, N lies on M_{a}N_{a} and M_{b}N_{b}; thus the three are indeed concurrent.

### Nine Point Circle

- Nine Point Circle: an Elementary Proof
- Feuerbach's Theorem
- Feuerbach's Theorem: a Proof
- Four 9-Point Circles in a Quadrilateral
- Four Triangles, One Circle
- Hart Circle
- Incidence in Feuerbach's Theorem
- Six Point Circle
- Nine Point Circle
- 6 to 9 Point Circle
- Six Concyclic Points II
- Bevan's Point and Theorem
- Another Property of the 9-Point Circle
- Concurrence of Ten Nine-Point Circles
- Garcia-Feuerbach Collinearity
- Nine Point Center in Square

|Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

71767405