Chapter 5, Problem 23CUR

To determine

To Complete: the statement $\angle ADC\cong \angle ?\cong \angle ?\cong \angle ?\cong ?$ such that $ABCD$ is a parallelogram, $\overrightarrow{AD}\cong \overrightarrow{AC}$ , $\overrightarrow{AE}\cong \overrightarrow{EC}$ , $\angle ADF\cong \angle CDF$ and $m\angle DAC=36$ ..

Answer to Problem 23CUR

  $\angle ABC,\angle ACD,\angle CDF\text{ and }\angle BAC$

Explanation of Solution

Given; Figure is shown as.

  

  $ABCD$ is a parallelogram, $\overrightarrow{AD}\cong \overrightarrow{AC}$ , $\overrightarrow{AE}\cong \overrightarrow{EC}$ , $\angle ADF\cong \angle CDF$ and $m\angle DAC=36$ .

Calculation:

  $m\angle BCA=m\angle DAC=36$ , alternate interior angles.

  $\angle ADC=\angle ACD\therefore \left(AD=AC\right)$

In $\Delta ADC$ .

  $\begin{array}{l}\angle DAC+\angle ADC+\angle ACD=180\\ 36+\angle ADC+\angle ADC=180\\ 2\angle ADC=144\\ \angle ACD=\angle ADC={72}^{\circ }\end{array}$

  $\angle ABC\cong \angle ADC=72\therefore \left(\text{Opposite angles of a parallelogram}\right)$

In $\Delta ABC$ .

  $\begin{array}{l}\angle ABC+\angle BCA+\angle BAC=180\\ 72+36+\angle BAC=180\\ \angle BAC=72\end{array}$

Therefore, $\angle BAC=72=\angle ADC$

In $\Delta CFD$ .

  $\begin{array}{l}\angle FCD+\angle CFD+\angle CDF=180\\ 72+\frac{1}{2}\angle ADC+\angle CDF=180\\ 72+\frac{1}{2}\times 72+\angle CDF=180\\ \angle CDF=72\end{array}$

Therefore, $\angle CDF=72=\angle ADC$

So, $\angle ADC=\angle ABC=\angle ACD=\angle CDF=\angle BAC=72$

Therefore, the answer is $\angle ABC,\angle ACD,\angle CDF\text{ and }\angle BAC$