Melody

Why are you answering when Ayden is a very obvious cheat?

Answering obvious cheats just gives this forum a really bad reputation.

Why would I delete it.

And why would anyone answer it?

You are obviously cheating. 

I hope you get caught and thrown out of your education institution.

Thanks for the info guest.    It is always good to get interesting background information.

Jugoslav was still asked to prove it though. 

He couldn't just say. It is true because Ptolomy said so.

That is what I thought.

I am sure these pricey courses are fantastic for some students.

But for most students they would not be. 

The work would have to be presented at the ideal level and the student would need to be very diligent.

There would also need to be a good help system.  Even if it is only talking to and learning with/from other students. 

To learn anything new you need to be able to ask your own questions and learn directly from other people.

This is my experience anyway.

Otherwise it would be like an unused gym membership.  A total waste of money.

Mmm 

You are using terms and formula that I have not seen before.

But yes, this is probably another way to skin the cat.  (sorry feline lovers - it is just an expression)

Hi Oofpirate.

What a nice introduction.  

It will be good to get to know you .   

Thanks Dragen   

I really do think that there must be a better way to tackle this problem though.

like this

\(BD^2=144+16-96cos\angle BCD\\ BD^2=160-96cos\angle BCD\\~\\ but\;\;also\;\; BD^2=121+100-220cos \angle BAD\\ BD^2=221-220cos \angle BAD\\ BD^2=221+220cos \angle BCD\qquad \text{opposite angles in cyclic quads are supplementary}\\~\\ so\\ 160-96cos\angle BCD=221+220cos \angle BCD\\ -316cos\angle BCD=61\\ cos\angle BCD=\frac{-61}{316}\\ so\\ BD^2=160-96*\frac{-61}{316}\\ BD^2=\frac{14104}{79}\\ BD=\sqrt{\frac{14104}{79}}\\ \)

Now do the same for AC

etc

LaTex:

BD^2=144+16-96cos\angle BCD\\
BD^2=160-96cos\angle BCD\\~\\
but\;\;also\;\;
BD^2=121+100-220cos \angle BAD\\
BD^2=221-220cos \angle BAD\\
BD^2=221+220cos \angle BCD\qquad \text{opposite angles in cyclic quads are supplementary}\\~\\
so\\
160-96cos\angle BCD=221+220cos \angle BCD\\
-316cos\angle BCD=61\\
cos\angle BCD=\frac{-61}{316}\\
so\\
BD^2=160-96*\frac{-61}{316}\\
BD^2=\frac{14104}{79}\\
BD=\sqrt{\frac{14104}{79}}\\