Euclidean Geometry

• Nov 14th 2010, 08:06 PM
Pythagonacci
Euclidean Geometry
Attachment 19710The attachment consists of 2 exercises from my last week's homework that I attempted to prove using numerous strategies, but never found a good one.

I searched for a while for the characteristics of a cyclic quadrilateral to use for the set up of the first proof, but never got even that far, and thus didn't get very far at all (I know...No S*^%!). As for the second exercise, no axiom, theorem, definition, or combination seemed very strong to prove the part i proposition. That left part iii pretty hopeless.

Thanks!
• Nov 15th 2010, 08:08 AM
emakarov
According to PlanetMath, a necessary and sufficient condition for a quadrilateral to be cyclic, is that the sum of a pair of opposite angles be equal to 180 degrees.

Concerning part 1, let F be a point on m left of C. I believe there is a theorem that angle FCA = angle ABC (if you need help finding it, let us know). We have angle ADE = 180 - CDE = 180 - FCA. Thus, ADE + ABE = 180. Similarly, it's easy to show that DAB + DEB = 180.
• Nov 15th 2010, 12:39 PM
Pythagonacci
Curious about the theorem
Thank you very much for the help emakarov! Please forgive me if I'm being dense, but the claim that angle FCA = angle ABC is difficult for me to accept given that F is an arbitrary point on m, only constrained such that it lies to the left of C. Because angle ABC is fixed once constructed, my wee brain cannot find a satisfactory way to believe the proposition.

I searched a bit for the theorem you mentioned, but didn't get far because I really don't know what I might search for, i.e. what to guess that the theorem might be called.

Notwithstanding all that, thanks for your help. (Bow)
• Nov 15th 2010, 01:06 PM
Plato
Put \$\displaystyle F\$ on \$\displaystyle m\$ so that \$\displaystyle A~\&~F\$ are on the same side of \$\displaystyle \overleftrightarrow {BC}\$

Now angles \$\displaystyle \angle ABC\;\& \,\angle FCA\$ are congruent, they intercept the same arc.

Angles \$\displaystyle \angle CDE\;\& \,\angle FCA\$ are alternant interiors so are congruent.

Angles \$\displaystyle \angle CDE\;\& \,\angle EDA\$ are supplementary.

Therefore angles \$\displaystyle \angle ABE\;\& \,\angle EDA\$ are supplementary.

That proves the quadrilateral is cyclic.
• Nov 15th 2010, 01:37 PM
emakarov
Quote:

the claim that angle FCA = angle ABC is difficult for me to accept given that F is an arbitrary point on m
Since F is on m and left of C, moving it does not change the angle FCA!

As Plato said, \$\displaystyle \angle ABC\$ and \$\displaystyle \angle FCA\$ are equal because they are half the arc AC. I found this explanation on cut-the-knot.org. You can also search for "inscribed angle".
• Nov 19th 2010, 12:00 AM
Pythagonacci
Thanks! Your responses led me to finally getting the idea through my rather thick skull.

I'm about to post another question that I find really interesting. After a good bit of consideration, I haven't come up with a solution for. Hope you check it out.

Thanks again!