Given:$\mathbf{ABCD}$is a parallelogram;$\overline{\mathrm{AD}}\cong \overline{\mathrm{AC}}$;$\overline{\mathrm{AE}}\cong \overline{\mathrm{EC}}$;$\angle \mathbf{ADF}\cong \angle \mathbf{CDF}$;$\mathbf{m}\angle \mathbf{DAC}\mathbf{=}\mathbf{36}$.

Complete each statement about the diagram.

$△\mathbf{ADF}$is a(n)$\underset{\_}{?}$ triangle.

The given diagram is:

It is being given that $\overline{AD}\cong \overline{AC}$.

The angles opposite to the equal sides are also equal.

As,$\overline{AD}\cong \overline{AC}$ , therefore, the angles opposite to the sidesAD and ACare also equal.

That implies,$\angle ACD=\angle ADC$ .

As the sum of the angles of the triangle is$180°$ .

Therefore, the sum of the angles of the triangle$ACD$ is$180°$ .

Therefore, it can be obtained that:

$\begin{array}{c}\angle DAC+\angle ACD+\angle ADC=180°\\ 36°+\angle ADC+\angle ADC=180°\\ 2\angle ADC=180°-36°\\ 2\angle ADC=144°\\ \angle ADC=\frac{144°}{2}\\ \angle ADC=72°\end{array}$

Therefore, the measure of the angle$\angle ADC$ is$72°$ .

It is also being given that$\angle ADF\cong \angle CDF$.

Therefore,$\angle ADF=\angle CDF$ .

From the diagram it can be noticed that:

$\angle ADC=\angle ADF+\angle CDF$.

Therefore, it can be obtained that:

$\begin{array}{c}\angle ADC=\angle ADF+\angle CDF\\ 72°=\angle ADF+\angle ADF\\ 72°=2\angle ADF\\ \frac{72°}{2}=\angle ADF\\ 36°=\angle ADF\end{array}$

Therefore, the measure of the angle$\angle ADF$ is $36°$.

It is given that$m\angle DAC=36$ .

From the given diagram it can be seen that$m\angle DAC=m\angle DAF=36$ .

Therefore, it can be seen that$\angle ADF=\angle DAF=36°$

In the triangle $△ADF$, it can be seen that$\angle ADF=\angle DAF=36°$.

Therefore, the two angles of the triangle are equal and so the triangle is an isosceles triangle.

$△ADF$is an isosceles triangle.