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.
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
(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 $AE$