Given,
[tex]\bar{AB}\cong\bar{AE}[/tex]
Let ∠AB=∠AE=x
The sum of angle in a triangle=180°
Hence,
[tex]x+x+24^0=180^0[/tex][tex]\begin{gathered} 2x+24^0=180^0 \\ 2x=180^0-24^0 \\ 2x=156^0 \\ x=\frac{156^0}{2} \\ x=78^0 \end{gathered}[/tex]
Let us get ∠ABC,
[tex]\begin{gathered} \text{The sum of angles in a straight line=180}^0 \\ \angle ABC=180^0-x=180^0-78^0 \\ \angle ABC=102^0 \end{gathered}[/tex]
Let us get ∠BAD,
∠BAD is perpendicular to ∠EAB
[tex]\begin{gathered} \angle BAD=90^0-24^0=66^0 \\ \angle BAD=66^0 \end{gathered}[/tex]
Let us now get ∠ADC,
The sum of angles in a quadrilateral is 360°.
[tex]\angle ADC=360^0-(\angle ABC+\angle BAD+\angle BCD)[/tex][tex]\begin{gathered} \angle ADC=360^0-(102^0+66^0+65^0) \\ \angle ADC=360^0-233^0 \\ \angle ADC=127^0 \end{gathered}[/tex]
To solve for ∠CDF,
The sum of angles in a straight line is 180°.
[tex]\begin{gathered} \angle CDF=180^0-\angle ADC \\ \angle CDF=180^0-127^0 \\ \angle CDF=53^0 \end{gathered}[/tex]
Hence, the m∠CDF is 53°.
