Triangle in a circle

Geometry Level 2

Consider a circle with AB as its diameter. C is a point on the circumference of the circle and D is the foot of the altitude from C onto AB such that AD = C D 2 \frac{CD}{2} . If A D B D \frac{AD}{BD} = a b \frac{a}{b} where a and b are coprime positive integers, then a+b=?


The answer is 5.

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Noel Lo by Noel Lo , 24, Singapore

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1 solution

Noel Lo
Apr 23, 2015

Angle ACB is a right angle since it is an angle subtended by a semicircle. Since triangle ACD is similar to CBD, by comparing angles, A D C D \frac{AD}{CD} = C D B D \frac{CD}{BD} = 1 2 \frac{1}{2} . So CD = 1 2 \frac{1}{2} * BD. Now, we have AD = 1 2 ∗ B D 2 \frac{\frac{1}{2} * BD}{2} = B D 4 \frac{BD}{4} so A D B D \frac{AD}{BD} = 1 4 \frac{1}{4} . Now a+b = 1+4 = 5 \boxed{5} .

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Alternatively, with A D C D \frac{AD}{CD} = C D B D \frac{CD}{BD} , A D B D \frac{AD}{BD} = A D C D \frac{AD}{CD} * C D B D \frac{CD}{BD} = ( A D C D ) 2 (\frac{AD}{CD})^2 = ( 1 2 ) 2 (\frac{1}{2})^2 = 1 4 \frac{1}{4}

Noel Lo - 6 years, 1 month ago

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