We are given a circle with intersecting chords $AC$ and $BD$. The measure of arc $AB$ is $41^\circ$ and the measure of arc $CD$ is $59^\circ$. We are asked to find the measure of angle $CED$.

1. Problem Description

We are given a circle with intersecting chords $AC$ and $BD$. The measure of arc $AB$ is $41^\circ$ and the measure of arc $CD$ is $59^\circ$. We are asked to find the measure of angle $CED$.

2. Solution Steps

The measure of an angle formed by two chords intersecting inside a circle is equal to one-half the sum of the intercepted arcs. In this case, angle $CED$ is formed by the intersection of chords $AC$ and $BD$. The intercepted arcs are $AB$ and $CD$. Therefore,

$m\angle CED = \frac{1}{2}(m\widehat{AB} + m\widehat{CD})$

We are given $m\widehat{AB} = 41^\circ$ and $m\widehat{CD} = 59^\circ$. Substituting these values into the formula:

$m\angle CED = \frac{1}{2}(41^\circ + 59^\circ)$

$m\angle CED = \frac{1}{2}(100^\circ)$

$m\angle CED = 50^\circ$

3. Final Answer

$m \angle CED = 50^\circ$