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In the given circle |AB| = 10 Units and AB || CD

AB is the diameter of the circle.

What is the length of the chord ED?

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Is $AB$ supposed to be a diameter of the circle? –  Robert Israel Sep 14 '12 at 7:37
    
Yes AB is the diameter of the circle. –  Sumit Bhowmick Sep 14 '12 at 7:38
    
I would be intrigued to know if it is possible without relying on trigonometry for the final calculation. –  Arthur Sep 14 '12 at 10:29

3 Answers 3

up vote 3 down vote accepted

By inscribed angle of a cirle, $\angle CBE = \angle CDE = 15^\circ$. That means $\angle ABE = 25^\circ$. Since $AB$ is a diameter, $\angle AEB$ is right, and thus we conclude that $\angle BAE = 90^\circ - 25^\circ = 65^\circ$.

Since $\angle BAD = 10^\circ$ by symmetry, we have that $\angle DAE = 65^\circ - 10^\circ = 55^\circ$. If we let $O$ be the center of the circle, we know then that $\angle DOE = 2\cdot\angle DAE = 110^\circ$. Your unknown side is now the last side of an isosceles triangle with two sides $5$ long, since they are radii in the circle, and with angles $110^\circ$, $35^\circ$ and $35^\circ$. That should be easer to calculate.

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Common case $\angle ABC = \alpha, \angle CDE = \beta$: enter image description here

Let $DF || BE \Rightarrow |ED|=|BF|$.

$\angle CBE = \angle EDC = \beta$ - same chord. $\angle EBF = \angle BFD= \angle DCB= \angle ABC =\alpha$.

As $\angle AFB=\frac{\pi}{2} \Rightarrow |FB|=|AB|\cdot \textbf{cos} \angle ABF$.

So $|ED|=|AB|\cdot \textbf{cos} (2\alpha+\beta)$.

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Sorry to bother but what tool do you use to draw images like the one you posted? –  Gabber Sep 14 '12 at 17:26
    
GeoGebra. geogebra.org –  Mike Sep 14 '12 at 17:51
    
Many many thanks –  Gabber Sep 14 '12 at 17:53
    
@Mike: Manu many thanks –  Sumit Bhowmick Sep 28 '12 at 4:57

enter image description here

O is center of circle. and You must know that total angle of circle is 360 degree.

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