# Simple Dissection of an Acute Triangle

#### 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 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.

  Attach:Dissection2.png Δ