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: :$2 \angle CED + \paren {180^\circ - \angle EDF} + \paren {180^\circ - \angle CAB} = 360^\circ$ 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. {{qed|lemma}} 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. Let $\triangle ABC$ be an obtuse triangle such that either $\angle ACB - \angle CAB > 90^\circ$ or $\angle ACB - \angle CBA > 90^\circ$. {{WLOG}}, suppose $\angle ACB - \angle CAB > 90^\circ$ and that $CAB \le CBA$. Let $P$ be constructed on $AB$ such that $AP = AC$. Then $\triangle ACP$ is an isosceles triangle with an acute apex. Thus from Isosceles Triangle with Acute Apex is Acute Triangle, $\triangle ACP$ is an acute triangle. :500px We have that: :$\angle CPB + \angle CPA = 180^\circ$ and: :$2 \angle CPA + \angle CAB = 180^\circ$ Thus: :$\angle CPB = 90^\circ + \dfrac {\angle CAB} 2$ {{begin-eqn}} {{eqn | l = \angle CPB       | r = 90^\circ + \dfrac {\angle CAB} 2       | c =  }} {{eqn | ll= \leadsto       | l = \angle CPB - \angle CBA       | r = 90^\circ + \dfrac {\angle CAB} 2 - \angle CBA       | c =  }} {{eqn | o = <       | r = 90^\circ       | c = as $\angle CBA \ge CAB$ }} {{end-eqn}} Also: {{begin-eqn}} {{eqn | l = \angle PCB + \angle CPB + \angle CBA       | r = 180^\circ       | c =  }} {{eqn | ll= \leadsto       | l = \angle PCB       | r = 90^\circ - \dfrac {\angle CAB} 2 - \angle CBA       | c =  }} {{eqn | ll= \leadsto       | l = \angle CPB - \angle PCB       | r = 90^\circ + \dfrac {\angle CAB} 2 - \paren {90^\circ - \dfrac {\angle CAB} 2 - \angle CBA}       | c =  }} {{eqn | r = \angle CAB + \angle CBA       | c =  }} {{eqn | o = <       | r = 90^\circ       | c = as $\angle ABC + \angle BAC + \angle ACB = 180^\circ$ and $\angle ACB > 90^\circ$ }} {{end-eqn}} Thus we have that $\triangle CPB$ fulfils the criteria to have a dissected into $7$ acute triangles. Together with acute triangle $\triangle ACP$, it is seen that $\triangle ABC$ has a dissected into $8$ acute triangles. {{qed|lemma}} It remains to be shown that at least $7$ pieces are always needed. Let $\mathcal X$ be a minimal dissection of $\triangle ABC$ into $n$ acute triangles. Suppose one of those dissecting lines be drawn from any vertex of $\triangle ABC$ to the opposite side of $\triangle ABC$. Then this divides $\triangle ABC$ into two triangles at least one of which is obtuse. This obtuse triangle will itself require $n$ pieces. Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature. Suppose one of its dissecting lines be drawn from $AC$ to $BC$. This cuts off an obtuse triangle with $\angle ACB$ as a vertex. This again needs $n$ pieces. Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature. Let lines be drawn from $D$ on $AC$ and $E$ on $BC$ to a single point $P$ on $AB$. Then either or both of $\triangle ADP$ or $\triangle BEP$ are themselves obtuse triangles. This again needs $n$ pieces. Hence $\triangle ABC$ needs at least $n + 1$ pieces, contradicting its minimal nature. The remaining option is for two lines from $AC$ to $AB$ and $BC$ to $AB$ at two separate points such that the parts cut off are acute triangles, leaving a pentagonal shape in the middle. From Dissection of Pentagon into 5 Acute Triangles, this needs at least $5$ acute triangles to dissect it. Hence the result. {{qed}}