*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.

Attach:Dissection.png Δ| **Figure 1**

**Proof:**If AH_{a}and BH_{b}are the altitudes to sides BC and AC, respectively, M_{c}is the midpoint of AB, it is well known that M_{c}H_{a}= M_{c}H_{b}= c/2. Accordingly, we have:Δ AM_{c}H_{b}is isosceles with base angles \alpha , and BM_{c}H_{a}is also isosceles triangle with base angles \beta . Δ H_{b}M_{c}H_{a}is isosceles too, let \gamma ' be the base angles of Δ H_{b}M_{c}H_{a}. Let's find the angles of the quadrilateral ABH_{a}H_{b}. Now,

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

- Let's observe now angles at the points H
_{a}and H_{b}. ∠H_{b}H_{a}C is equal to \alpha , because ∠H_{b}H_{a}C + \beta + \gamma = 180° (the straight angle CH_{a}B). Analogously, ∠H_{a}H_{b}C is equal to \beta .Moreover, the angle between lines AB and H_{a}H_{b}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 H_{a}) angle and two interior angles (at the points P and B) of the triangle PBH_{a}. In conclusion: X = \alpha - \beta .

Another way to obtain the same result is to observe that ∠AM_{c}H_{b} = 180° - 2\alpha while ∠BM_{c}H_{a} = 180° - 2\beta, implying that ∠H_{a}M_{c}H_{b} = 180° - 2\gamma. Since ΔH_{a}M_{c}H_{b} is isosceles, it's base angles are found to equal \gamma. Now it is immediate that at H_{a} and H_{b} we have triples of the angles \alpha , \beta , \gamma , as claimed.

#### Addendum:

- In ΔABC, ΔH
_{a}H_{b}H_{c}is known as the orthic triangle. It has an interesting property that its angle bisectors serve in fact as altitudes of ΔABC.

Attach:Dissection2.png Δ

**Proof:**As we know, ∠H_{b}H_{a}C = \alpha ( see above ), because ΔABC is similar to ΔH_{a}H_{b}C. Since ΔACB is similar to ΔH_{a}H_{c}B, ∠H_{C}H_{a}B = \alpha and AH_{a}is perpendicular to line BC, we conclude that line AH_{a}is the angle bisector ( ∠AH_{a}H_{b}= ∠AH_{a}H_{c}= 90° - \alpha ) of the angle at the point H_{a}of the orthic triangle H_{a}H_{b}H_{c}. The angles at the points H_{b}and H_{c}we can treat similarly. It is clear now, that orthocenter of the triangle ABC is simultaneously incenter of the triangle H_{a}H_{b}H_{c}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.