Welcome to the forum.
This is a long thread and there's at least one other on this question. The answer to your question is in there somewhere; you just need to go back to the early posts. I recommend you use my labelling rather than using M and N. Then you'll find the answer.
Bob
]]>There are no posts about movie, I see.
A space odyssey, first page
]]>If ANB = 90, the midpoint of the hypotenuse, M, will also be the circumcenter of ANB, therefore AM=BM=NM. I don't know how to prove that ANB is a right triangle.
I also used some arc lengths and tangents to prove that angle BDP is congruent to ABP
This forum is kind of informal and friendly. Ask a question and help comes, it is that easy.
There really was only a bit about that movie, this thread is stuffed, I mean stuffed with geometry.
]]>How do you know that AB=LM without using geogebra?
AB & LM are opposite sides of rectangle ABML (from post #24):
Proof that H is CD's midpoint:
(i) ABML is a rectangle (tangent perpendicular to radius theorem [see 'Tangent Angle' at bottom of page here] for points A & B; AB is parallel to CD [see post #1]; AE extends to L)
Good luck, Bob!
Yes, I thought there should be an easier way, too...as there was with this one, with ΔPEB being 1/3rd of ΔPCA.
Here's another diagram, with another property revealed by Geogebra: ie, CL = LP = HM. But again, I don't know how to use that info.
This time I've marked some of the equal-length lines.
This was the digram I was referring to.
]]>This wasn't my proof, but here's how I know this to be true. The circle centred on G defines the point H. AHB is 90 at it is the angle at the circumference subtended by the diameter AB **. AB is parallel to CD so angle AHB = HBD (use x + y = 90).
Bob
** http://www.mathisfunforum.com/viewtopic.php?id=17799 post #6
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