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The angle $$\alpha$$ is formed by the intersection of chords AC and BD. The measure of such an angle is half the sum of the measures of the intercepted arcs. The intercepted arcs are arc AB and arc CD. The problem states that arc AC = 62 degrees and arc BD = 16 degrees. However, the angle $$\alpha$$ intercepts arc BC and arc AD. We are given arc AC and arc BD. The angle $$\alpha$$ is vertically opposite to the angle formed by the intersection of AC and BD. Let's assume the question is asking for the angle formed by the intersection of chords AB and CD, which is labeled as $$\alpha$$. This angle intercepts arcs AC and BD. Therefore, $$\alpha = (arc AC + arc BD) / 2$$. Given arc AC = 62 degrees and arc BD = 16 degrees. $$\alpha = (62 + 16) / 2 = 78 / 2 = 39$$ degrees. If $$\alpha$$ refers to the angle that intercepts arc BC, then we need more information. Based on the diagram, $$\alpha$$ is shown as an angle formed by the intersection of chords AB and CD. Thus, $$\alpha = (arc AC + arc BD) / 2 = (62 + 16) / 2 = 39$$ degrees.