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The chord CD of a circle center O is perpendicular to the diameter AB. The chord AE goes through the midpoint of the radius OC. Prove that the chord DE goes through the midpoint BC.

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This question does not show any research effort –  The Chaz 2.0 Oct 2 '11 at 5:09
    
2000clicks.com/MathHelp/… –  Bhargav Oct 2 '11 at 5:59
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Is it necessary that all questions other than research to be marked as Homework?? A friend of mine asked me to solve this Question. –  Ramana Venkata Oct 2 '11 at 8:26
    
A inelegant way to prove this would be to resort to Cartesian coordinates. –  user17762 Oct 24 '11 at 18:04
    
@SivaramAmbikasaran But it would be very difficult to solve the equations although I agree that it can proved –  Ramana Venkata Oct 24 '11 at 18:15

2 Answers 2

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(Sketch.) Triangles OCA and BCD are similar. The similarity transformation sending OCA to BCD (a rotation with centre C by angle ACD, followed by some scaling) sends the median of OCA from vertex A (that is, the line AE) to the median $m$ of BCD from D. Since $m$ is the image of AE under the similarity transformation, the angle between m and AE is angle ACD, which equals angle AED. Thus DE is $m$.

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It does not have to be the midpoint of $CO$; in general given $M$ on $OC$ and $N$ as the intersection of $BC$ and $DE$, $MN \parallel AB$. Here is my proof:

Let $K$ be the intersection of $BC$ and $AE$
Then cross-ratios $(C,B;K,N) = (C,B;A,D)$ give $\frac{CK}{KB} / \frac{CN}{NB} = \frac{CA}{AB} / \frac{CD}{DB} = \frac{1}{2}$ by similar right-angled triangles
By Menelaus' Theorem, $-1 = \frac{CK}{KB} \frac{BA}{AO} \frac{OM}{MC} = ( \frac{1}{2} \frac{CN}{NB} ) (-2) \frac{OM}{MC}$
Thus $\frac{CN}{NB} = \frac{CM}{MO}$
(QED)

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