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.
Problem on circles- Plane Geometry
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0@SivaramAmbikasaran But it would be very difficult to solve the equations although I agree that it can proved – 2011-10-24
2 Answers
(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 $m$.
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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0@BillDubuque: Sure. I am definitely open to that as well. And thanks for reaching out. Perhaps we should talk more about other things, such as logic and AHK, and you're welcome to [the logic chat-room](https://chat.stackexchange.com/rooms/44058/logic) that I started, and/or [SBA's realm](https://chat.stackexchange.com/rooms/51337/this-is-the-realm-of-simply-beautiful-art) for random stuff. =) – 2018-11-04