Question

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.

.$\angle \mathbf{ADC}\cong \angle \underset{\_}{?}\cong \angle \underset{\_}{?}\cong \angle \underset{\_}{?}\cong \angle \underset{\_}{?}$ .

Short answer

$\angle ADC\cong \angle \mathbf{ABC}\cong \angle \mathbf{ACD}\cong \angle \mathbf{CFD}\cong \angle \mathbf{BAC}$

Step-by-step solution

- Observe the given diagram.

The given diagram is:

- Description of step.

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$72°$ .

Therefore,$\angle ACD=\angle ADC=72°$ .

Therefore,$\angle FCD=\angle ACD=72°$ .

- Description of step.

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 $36°$.

Therefore, $\angle ADF=\angle CDF=36°$.

- Description of step.

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

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

Therefore, it can be obtained that:

$\begin{array}{c}\angle CFD+\angle CDF+\angle FCD=180°\\ \angle CFD+36°+72°=180°\\ \angle CFD+108°=180°\\ \angle CFD=180°-108°\\ \angle CFD=72°\end{array}$

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

- Description of step.

In the parallelogram, both pairs of opposite angles are congruent.

Therefore,$\angle ABC=\angle ADC=72°$

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

In the parallelogram, both pairs of opposite sides are parallel.

As,$AB\parallel CD$ and ACis a transversal.

Therefore, it can be seen that the angles $\angle BAC$and $\angle ACD$are the alternate interior angles.

Therefore,$\angle BAC=\angle ACD=72°$

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

- Write the complete statement.

$\angle ADC\cong \angle ABC\cong \angle ACD\cong \angle CFD\cong \angle BAC$