0
$\begingroup$

$O$ is the center of a circle with $AB$ diameter. If $OD$ is perpendicular to $AB$ and meets chord $AC$ or $AC$ produced at $D$, then prove that circle $AOD$ is equal to the circle through $O,B,D,C$.

The triangles $AOD$ and $ACB$ are similar. I've tried using similarity for the above theorem but we need to show that the two radiuses are equal. Since, $AOD$ is a right-angled triangle, its circumcenter lies on the hypotenuse.

I am not able to proceed further. Please help.

Thanks

$\endgroup$
  • 1
    $\begingroup$ DA (the diameter of circle OAD) is symmetrically equal to DB( the diameter of circle ODB). $\endgroup$ – Mick Oct 4 '16 at 8:12
  • $\begingroup$ You should ponder to add a diagram to this kind of questions, otherwise it may be hard to follow . $\endgroup$ – DonAntonio Oct 4 '16 at 10:06
  • $\begingroup$ @Mick Thanks! $$$$ $\endgroup$ – TheRandomGuy Oct 4 '16 at 10:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.