jmc

Clearly, triangle $ABC$ is isosceles. This is the first. We know $\angle ABC=\angle ACB=72^{\circ }$, which tells us that $\angle BAC=180^{\circ }-72^{\circ }-72^{\circ }=36^{\circ }$ . Since segment $BD$ bisects angle $ABC$, the measure of angle $ABD$ is $72^{\circ }/2=36^{\circ }$. Thus, $\angle BAD=\angle ABD$ and $△ABD$ is isosceles.

Since $△ABD$ is isosceles, we see that $m\angle ADB=180^{\circ }-36^{\circ }-36^{\circ }=108^{\circ }$. Thus, $\angle BDC=180^{\circ }-108^{\circ }=72^{\circ }$. Looking at triangle $BDC$, we already know that $\angle DCB=72^{\circ }=\angle BDC$ degrees, so this triangle is isosceles.

Next, we use the fact that $DE$ is parallel to $AB$. Segment $BD$ is a transversal, so the alternate interior angles $ABD$ and $BDE$ are congruent. Thus, $m\angle ABD=m\angle BDE=36^{\circ }$. We already knew that $m\angle DBE=36^{\circ }$ since $BD$ bisects $\angle ABC$. Thus, the triangle $BDE$ is isosceles.

Looking at angle $EDF$, we can see that $m\angle EDF=180^{\circ }-m\angle BDA-m\angle BDE=180^{\circ }-108^{\circ }-36^{\circ }=36^{\circ }$. We also know that $EF$ is parallel to $BD$, and so the alternate interior angles $\angle BDE$ and $\angle FED$ are congruent. Thus, $m\angle FED=36^{\circ }$ and triangle $DEF$ is isosceles.

We have nearly found them all. We can compute that $\angle EFD=180^{\circ }-36^{\circ }-36^{\circ }=108^{\circ }$, and so $\angle EFC=180^{\circ }-108^{\circ }=72^{\circ }$ degrees. From the very beginning, we know that $m\angle ACB=72^{\circ }$, so $△FEC$ is isosceles. This makes $m\angle FEC=180^{\circ }-72^{\circ }-72^{\circ }=36^{\circ }$ degrees, and so $m\angle DEC=36^{\circ }+36^{\circ }=72^{\circ }$. So, our final isosceles triangle is $DEC$. We have found a total of $\boxed{7}$ isosceles triangles.