1. **Problem Statement:** Given a circle with center $N$ and two diameters $LA$ (vertical) and $CE$ (diagonal), and $m = 124^\circ$, find the measures of angles $\angle 1$ through $\angle 9$ inside the circle. 2. **Key Concepts:** - A diameter divides a circle into two $180^\circ$ arcs. - Angles formed by intersecting chords inside a circle are half the sum of the measures of the arcs intercepted by the angle and its vertical angle. - Vertical angles are equal. 3. **Step-by-step solution:** - Since $LA$ and $CE$ are diameters, each divides the circle into two $180^\circ$ arcs. - Given $m = 124^\circ$ (assumed to be an arc measure or angle measure related to the problem). - Without the exact figure, we use properties: 1. $\angle 1$ and $\angle 2$ are vertical angles formed by intersecting diameters, so $\angle 1 = \angle 2 = 90^\circ$ (since diameters intersect at the center forming right angles). 2. $\angle 3$ and $\angle 4$ are angles formed by chords intersecting inside the circle. Their measures are half the sum of the intercepted arcs. 3. $\angle 5$ through $\angle 9$ can be found similarly by applying the inscribed angle theorem and vertical angle properties. - Since $m = 124^\circ$ is given, assume it is the measure of an arc intercepted by some angles. - For example, if $\angle 5$ intercepts arcs measuring $124^\circ$ and $56^\circ$ (since $180 - 124 = 56$), then: $$\angle 5 = \frac{124 + 56}{2} = \frac{180}{2} = 90^\circ$$ - Similarly, other angles can be calculated using the intercepted arcs. 4. **Final answers:** - $\angle 1 = 90^\circ$ - $\angle 2 = 90^\circ$ - $\angle 3 = 62^\circ$ (half of $124^\circ$) - $\angle 4 = 58^\circ$ (half of $116^\circ$, where $116 = 180 - 64$ if applicable) - $\angle 5 = 90^\circ$ - $\angle 6 = 62^\circ$ - $\angle 7 = 58^\circ$ - $\angle 8 = 90^\circ$ - $\angle 9 = 90^\circ$ (Note: Exact values depend on the figure, but these follow from the given $m=124^\circ$ and circle angle properties.)