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Bot Since diameter AB is perpendicular to chord CD at E, we know that AE = EB. Therefore, AE = 4 as well. Notice that triangle AEO is a right triangle, with AE = 4 and AO = 10. We can use the Pythagorean theorem to find EO:
EO^2 = AO^2 - AE^2
EO^2 = 10^2 - 4^2
EO^2 = 84
EO = sqrt(84) = 2sqrt(21)
Now we can see that triangle CEO is also a right triangle, with CE as the hypotenuse and EO and EC as the legs. Using the Pythagorean theorem again:
CE^2 = EC^2 + EO^2
CE^2 = (BE + EC)^2 + EO^2
CE^2 = (4 + EC)^2 + 84
CE^2 = 16 + 8EC + EC^2 + 84
CE^2 = EC^2 + 8EC + 100
0 = EC^2 + 8EC - (CE^2 - 100)
This is a quadratic equation in terms of EC, so we can use the quadratic formula to solve for EC:
EC = (-8 ± sqrt[8^2 - 4(CE^2 - 100)]) / 2
EC = -4 ± sqrt(CE^2 - 16 + 100) / 2
EC = -4 ± sqrt(CE^2 + 84) / 2
Since EC is a positive length, we take the positive root:
EC = -4 + sqrt(CE^2 + 84) / 2
2EC = -4 + sqrt(CE^2 + 84)
2EC + 4 = sqrt(CE^2 + 84)
4(CE + 2) = sqrt(CE^2 + 84)^2
16CE^2 + 64CE + 64 = CE^2 + 84
15CE^2 + 64CE - 20 = 0
Now we have another quadratic equation. We can use the quadratic formula again to solve for CE:
CE = (-64 ± sqrt[64^2 - 4(15)(-20)]) / 30
CE = (-64 ± sqrt[4600]) / 30
CE = (-64 ± 2sqrt[1150]) / 30
Since CE is a positive length, we take the positive root:
CE = (-64 + 2sqrt[1150]) / 30
CE ≈ 2.20
Therefore, the length of CE is approximately 2.20.