Vladimir Nikolin

Statement: Every acute triangle can be partitioned into three isosceles triangles (with the base angles equal to \alpha , \beta , \gamma ) and one triangle with the same angles (similar) as the original.

Figure 1
Proof: If AHa and BHb are the altitudes to sides BC and AC, respectively, Mc is the midpoint of AB, it is well known that McHa = McHb = c/2. Accordingly, we have:
Δ AMcHb is isosceles with base angles \alpha , and BMcHa is also isosceles triangle with base angles \beta . Δ HbMcHa is isosceles too, let \gamma ' be the base angles of Δ HbMcHa. Let's find the angles of the quadrilateral ABHaHb. Now,

2 \alpha + 2 \beta + 2 \gamma ' = 360 or \alpha + \beta + \gamma ' = 180, thus \gamma ' = \gamma .

Let's observe now angles at the points Ha and Hb. ∠HbHaC is equal to \alpha , because ∠HbHaC + \beta + \gamma = 180 (the straight angle CHaB). Analogously, ∠HaHbC is equal to \beta .
Moreover, the angle between lines AB and HaHb is equal to \alpha - \beta (If \alpha > \beta ). Why? So, let X be the required angle. By the Exterior Angle Theorem, the sum of the two remote interior angles equals the exterior angle. We have \beta + X = \alpha , as one exterior (at the point Ha) angle and two interior angles (at the points P and B) of the triangle PBHa. In conclusion: X = \alpha - \beta .

Another way to obtain the same result is to observe that ∠AMcHb = 180° - 2\alpha while ∠BMcHa = 180° - 2\beta, implying that ∠HaMcHb = 180° - 2\gamma. Since ΔHaMcHb is isosceles, it's base angles are found to equal \gamma. Now it is immediate that at Ha and Hb we have triples of the angles \alpha , \beta , \gamma , as claimed.


In ΔABC, ΔHaHbHc is known as the orthic triangle. It has an interesting property that its angle bisectors serve in fact as altitudes of ΔABC.
Proof: As we know, ∠HbHaC = \alpha ( see above ), because ΔABC is similar to ΔHaHbC. Since ΔACB is similar to ΔHaHcB, ∠HCHaB = \alpha and AHa is perpendicular to line BC, we conclude that line AHa is the angle bisector ( ∠AHaHb = ∠AHaHc = 90° - \alpha ) of the angle at the point Ha of the orthic triangle HaHbHc. The angles at the points Hb and Hc we can treat similarly. It is clear now, that orthocenter of the triangle ABC is simultaneously incenter of the triangle HaHbHc and vice versa. if ΔABC is obtuse then we can use the fact that if H is the orthocenter of ΔABC, then C is the orthocenter of ΔABH.