To solve the problem, let's analyze the geometry step-by-step.

Problem Overview:

We are tasked with finding the measure of $\angle CEF$, given:

1. $\overline{AE}$ and $\overline{BD}$ intersect at $C$.
2. $\overline{DE}$ is extended from $E$ to $F$.
3. $\overline{AC} \cong \overline{BC}$ (isosceles triangle $\triangle ABC$).
4. $m\angle A = 70^\circ$.
5. $m\angle D = 80^\circ$.

Step 1: Solve $\triangle ABC$

Since $\triangle ABC$ is isosceles with $\overline{AC} \cong \overline{BC}$, the base angles are congruent: $m\angle B = m\angle A = 70^\circ$ The third angle, $\angle ACB$, can be calculated as: $m\angle ACB = 180^\circ - m\angle A - m\angle B = 180^\circ - 70^\circ - 70^\circ = 40^\circ$

Step 2: Analyze $\triangle CDE$

Given $m\angle D = 80^\circ$, and since $\triangle CDE$ is a triangle: $m\angle C + m\angle D + m\angle E = 180^\circ$ Substitute $m\angle D = 80^\circ$ and $m\angle C = \angle ACB = 40^\circ$: $40^\circ + 80^\circ + m\angle E = 180^\circ$ $m\angle E = 60^\circ$

Step 3: Determine $\angle CEF$

$\overline{DE}$ is extended to $F$, making $\angle CEF$ an exterior angle of $\triangle CDE$. The exterior angle theorem states: $m\angle CEF = m\angle C + m\angle D$ Substitute $m\angle C = 40^\circ$ and $m\angle D = 80^\circ$: $m\angle CEF = 40^\circ + 80^\circ = 120^\circ$

Final Answer:

$m\angle CEF = 120^\circ$

Would you like additional clarification or a step-by-step explanation of any part?

Here are 5 related questions for further understanding:

1. How can the exterior angle theorem be applied to other triangles?
2. What makes a triangle isosceles, and how do we identify congruent angles?
3. What are the properties of exterior angles in polygons?
4. Can you verify this solution by breaking down the problem further?
5. How can extending a side of a triangle help in solving for unknown angles?

Tip: When working with isosceles triangles, always consider their base angles as equal—it simplifies many angle computations!