Obtuse Triangle Divided into Acute Triangles

Theorem
Let $T$ be an obtuse triangle.

Let $T$ be dissected into $n$ of acute triangles.

Then $n \ge 7$.

Construction
Let $\triangle ABC$ be an obtuse triangle such that:
 * $\angle ACB > 90^\circ$
 * $\angle ACB - \angle CAB < 90^\circ$
 * $\angle ACB - \angle CBA < 90^\circ$

Let $D$ be the incenter of $\triangle ABC$.

Let a circle be constructed whose center is $D$ and whose radius is $CD$.

The dissection into acute triangles is then performed as follows:


 * ObtuseTriangleAcuteDissection.png

This gives a dissection into $7$ pieces.

If the conditions on $A$ and $B$ are not satisfied, then it is possible to draw a line from $C$ to $AB$ to dissect $\triangle ABC$ into an acute triangle and an obtuse triangle which does satisfy the conditions.

This gives a dissection into $8$ pieces.

Proof
As $D$ is equidistant from $AC$, $CB$ and $BA$, it follows that $\angle CDH = \angle CDE = \angle FDG$.

As $CD = DE = DF = DG = DH$, it follows that each of $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are isosceles.

From Triangle Side-Angle-Side Equality, $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are congruent.

Hence it follows that:
 * $\angle ECD + \angle CED = \angle ACB$

and that both $\angle ECD$ and $\angle CED$ are acute.

As $\angle ACB$ is obtuse, it follows that $\angle ECD + \angle CED$ together are greater than a right angle.

From Sum of Angles of Triangle equals Two Right Angles, it follows that $\angle CDE$ is acute.

Thus $\triangle CDE$, $\triangle CDH$ and $\triangle FDG$ are all congruent acute triangles.

As $\angle ACB$ is obtuse, it follows that $\angle CAB$ and $\angle CBA$ are both acute.

As $AE = AF$ and $BG = BH$, it follows that $\triangle EAF$ and $\triangle GBH$ are both isosceles.

As $\angle AEF = \angle AFE$ and $\angle AEF + \angle AFE$ are less than $2$ right angles, each of $\angle AEF$ and $\angle AFE$ are acute.

Thus $\triangle EAF$ is an acute triangle.

By the same argument, $\triangle GBH$ is also an acute triangle.

Thus it has been established that for obtuse $\angle ACB$, five of the triangles into which $\triangle ABC$ has been dissected are acute.

The conditions under which the remaining triangles $\triangle DEF$ and $\triangle DGH$ are also acute triangle are still to be established.

By the same argument as for $\triangle CDE$ and $\triangle CDH$, we also have that $\triangle DEF$ and $\triangle DGH$ are isosceles.

By the same argument as for $\triangle EAF$ and $\triangle GBH$, it follows that $\angle DEF$ and $\angle DFE$ are both acute, as are $\angle DHG$ and $\angle DGH$.

We have that $\angle CED + \angle DEF + \angle AEF$ form $2$ right angles.

From Sum of Angles of Triangle equals Two Right Angles, and the fact that $\triangle AEF$ and $\triangle DEF$ are both isosceles:
 * $2 \angle AEF = 180^\circ - \angle CAB$
 * $2 \angle DEF = 180^\circ - \angle EDF$

and so:
 * $2 \angle CED + \left({180^\circ - \angle EDF}\right) + \left({180^\circ - \angle CAB}\right) = 360^\circ$

which, when simplified, gives:
 * $\angle EDF = 2 \angle CED - \angle CAB$

But:
 * $2 \angle CED = \angle ACB$

as established above.

Hence $\triangle EDF$ is acute exactly when:
 * $\angle ACB - \angle CAB < 90^\circ$

Similarly, $\triangle CDH$ is acute exactly when:
 * $\angle ACB - \angle CBA < 90^\circ$

Thus it has been established that when:
 * $\angle ACB > 90^\circ$
 * $\angle ACB - \angle CAB < 90^\circ$
 * $\angle ACB - \angle CBA < 90^\circ$

$\triangle ABC$ can be dissected into $7$ of acute triangles.

As seen above, when $\angle ACB - \angle CAB \ge 90^\circ$, or $\angle ACB - \angle CBA \ge 90^\circ$, the above construction does not work.

However, it will be demonstrated that there exists an $8$ piece dissection of such a triangle. .