Question

Find the measure of $\angle DCE$ in degrees.

Answer 1

In $△ABC$,
Given: AB = AC
Hence, $\angle ABC=\angle ACB$ (Isosceles triangle property)
Now, $\angle BAF=\angle ACB+\angle ABC$ (Exterior angle propery)
$128=2\angle ACB$
$\angle ACB={64}^{\circ }$
Thus, $\angle ACB=\angle ABC={64}^{\circ }$
Now, In $△CBD$,
Given, CD = BC
Thus, $\angle CDB=\angle CBD={64}^{\circ }$ (Isosceles triangle property)
Sum of angles of the triangle = 180
$\angle CDB+\angle CBD+\angle BCD=180$
$64+64+\angle BCD=180$
$\angle BCD={52}^{\circ }$
$\angle BCA=\angle BCD+\angle DCE$
$64=52+\angle DCE$
$\angle DCE={12}^{\circ }$