ABC and CDE are isosceles triangles. In \( \triangle ABC \), \( AB = AC \) so base angles are equal. In \( \triangle CDE \), \( CD = CE \) so its base angles are equal.<br><br>
BCEGH is a straight line (\(180^\circ\)). Use the marked angles in the figure, angle sum of a triangle, and angles on a straight line to find \( \angle CDE \).<br><br>
\( \angle CDE = 100^\circ \).<br><br>
Final Answer: 100&deg;