Question

in the diagram below, \\(\overline{bc} \cong \overline{ec}\\) and \\(\text{m}\angle abe = 110^{\circ}\\). find \\(\text{m}\angle fcd\\).

angle	reason
\\(\text{m}\angle \square = \square^{\circ}\\)	select a reason

(diagram: a straight line a-b-c-d, with b connected to e and c connected to f and e. \\(\angle abe = 110^{\circ}\\).)

you may assume lines that appear straight are straight, but the figure is not otherwise drawn to scale.

Explanation:

Step1: Find supplementary angle

$\angle ABE$ and $\angle EBD$ are supplementary, so $\text{m}\angle EBD = 180^\circ - 110^\circ = 70^\circ$

Step2: Identify isosceles triangle angles

Since $\overline{BC} \cong \overline{EC}$, $\triangle BCE$ is isosceles, so $\text{m}\angle BEC = \text{m}\angle EBD = 70^\circ$

Step3: Calculate triangle's third angle

Sum of angles in a triangle is $180^\circ$, so $\text{m}\angle BCE = 180^\circ - 70^\circ - 70^\circ = 40^\circ$

Step4: Find vertical angle

$\angle FCD$ and $\angle BCE$ are vertical angles, so $\text{m}\angle FCD = \text{m}\angle BCE = 40^\circ$

Answer:

$\text{m}\angle FCD = 40^\circ$

(For the table blank: $\boldsymbol{\angle FCD = 40^\circ}$, Reason: Vertical angles are congruent (or derived from isosceles triangle and supplementary angle properties))