## Elementary Geometry for College Students (6th Edition)

Since ACB is an inscribed angle and is thus equal to half of the measure of 180 degrees, it follows that ACB is a right angle. Thus, we know: $AC^2 + CB^2 = AB^2$ We now solve for AC and CB. We call the length of CM k and the length CN j. Thus, we find: $(2k)^2 + j^2 = 73$ And: $k^2 + (2j)^2 = 52$ Thus, we obtain: $4(52 - 4j^2) + j^2 = 73 \\ -15j^2 = -135 \\ j = 3$ This means: $k = \sqrt{52 - 4(9)} = 4$ Since k is half of AC and j is half of CB, we find: $AB = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$