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1 solution
Solution 1:
Let O be the center of the circle, then angle DOB = 2C => OBD = (180-2C)/2 = 90-C as triangle BOD is isocelese.
So ABD = ABO-OBD = 90-(90-C) = C Also ABC is isocelese, so A = C Hence ADB = B = 180-2C and AD = BD
Using cos rule for ABC, we have 15² = 2 * 10² - 2*10² cos B => -2 cos B = (225-200)/100 = 1/4
Using cos rule for ADB, we have AB² = AD²+BD²-2 AD BD cos ADB => 10² = 2BD² + BD²*(-2 cos B) => 100 = BD² (2+1/4) = 9BD²/4 => BD² = 4 * 100/9 => BD = 20/3
Solution 2:
Since AB = BC, so A = C. Since ABD (an angle formed by a tangent and a chord) and C (an inscribed angle) are both subtended by the chord DB, they are equal, i.e. A = C = ABD. Therefore DB = AD. Now we seek DB. Equating the power of the point A in two different ways, we have
AD*AC = AB^2,
AD*15 = 10^2,
AD = 20/3.
It follows that DB = AD = 20/3.