Geometry proof

There is no D on the pic. 

Is D and O the same perhaps?

Also what does "measure arc LB" mean ? is it an angle or an arc length?

What does m

Hi Melody

This business of the measure of an angle seems to be something that's crept in (from the USA ?) only recently, the last few years ?

The measure of an arc seems to be even more recent, I certainly don't recall coming across it until a few months ago, I think that it's the angle subtended at the circles centre by the arc.

Maybe one of the sites american contributers can enlighten us ?

Thanks guest 

That is what I thought too but I don't like the terminology much.

What was wrong with how we did it in the old days. Lol

I still don't know how to answer the question though :((

Call the angle LNB \(\displaystyle \theta\) (I resist using the word measure !), then LRB will be \(\displaystyle \theta/2.\)

Call the angle RND (RNO ?) \(\displaystyle \phi\), then RBD (RBO ?) will be \(\displaystyle \phi/2.\)

etc..

Thanks very much guest.  

ok I'll do the proof in full

Consider minor arc LB

This subtends the angle LRB at the circumference.  And subtends angle LNB at the centre

So     < LNB= 2*< LRB

 \(Let \quad \angle LRB=\theta \qquad so \qquad \angle LNB=2\theta\)

Consider minor arc RD

This subtends the < RBD at the circumference.  And subtends

So    

\(Let \quad \angle RBD=\alpha \qquad so \qquad \angle RND=2\alpha\)

\(Consider\quad \triangle RBE\\ \angle LRB=\angle RBE+\angle REB \qquad \text{(Exterior angle of a triangle = sum of opposite interior angles)}\\ \theta=\alpha+\angle REB\\ \angle REB=\theta-\alpha\\ \angle REB=\frac{1}{2} (2\theta-2\alpha)\\ \angle REB=\frac{1}{2} (\angle LNB-\angle RND)\\ \)