# Geometry Proof

• Feb 3rd 2011, 11:03 AM
BabyMilo
Geometry Proof
Attachment 20678
OB=arcOA=$\displaystyle \theta$

Prove that OP= 1-cos$\displaystyle \theta$

Deduce that OD = $\displaystyle \frac{\theta(1-cos\theta)}{\theta-sin\theta}$

thanks.
• Feb 3rd 2011, 11:55 AM
emakarov
is it known that the circle's center lies on the line OP? Is C the center of the circle? If yes, it's pretty obvious that $\displaystyle CP = \cos\theta$.

For the second equality, let A' be the projection of A on OB. Use the fact that the triangle BA'A is similar to the triangle BOD.
• Feb 3rd 2011, 12:07 PM
BabyMilo
yes to your questions, but why cp=cos theta
i cant see it, im stupid :)
and can you explain the 2nd part in more details.

thanks.
• Feb 3rd 2011, 12:26 PM
emakarov
Quote:

Originally Posted by BabyMilo
why cp=cos theta

Because $\displaystyle \angle OCA=\theta$. The fact that $\displaystyle CP=\cos\theta$ follows from the definition of cosine.

Quote:

and can you explain the 2nd part in more details.
Draw a horizontal line through A and call its intersection with the line OB by A'. Then the triangle BA'A is similar to the triangle BOD because all three angles are equal. Therefore the ratio of A'A = OP to OD is the same as the raio of BA' to BO.