Is my proof of Simon's line theorem complete?

[disregardthat]

I have written a proof of the simons line theorem. I would appreciate if someone would tell me if it's complete.

We have \Delta ABC inscrubed in a circle, and P is an arbitrary point that does not coincide with either of the corners of the triangle. (See attached figure).

Now draw perpendiculars from P to [AB], [AC] and [BC] which intersect with the lines in U, \ Y and Z respectively.

Let E be the point of intersection between [PU] and [AY].
Let T be the point of intersection between [AC] and [BP].

Let \angle BZP=\alpha and \angle APU=\beta.

Now as \angle BZP and \angle BUP are supplementary, BZPU is a cyclic quadrilateral \Rightarrow \angle PUZ = \alpha.

Now as \angle PAC subtend over the same arc as \angle PBZ, \ \angle PAC = \alpha \Rightarrow \angle UAR = 90 - (\alpha +\beta) \Rightarrow \angle ARU = \alpha + \beta \Rightarrow \angle PET = \alpha +\beta.

Let \angle UZB = \gamma \Rightarrow \angle UPB = \gamma.

As \angle CYP and \angle CZP are supplementary, CZPY is a cyclic quadrilateral \Rightarrow \angle CPY = \gamma.

As \angle PAY =\angle PUY, APYU is a cyclic quadrilateral \Rightarrow \angle UYA = \beta.

Now \angle ETP =180-(\alpha +\beta + \gamma) \Rightarrow \angle PTY = \alpha +\beta +\gamma \Rightarrow \angle TPY=90-(\alpha+\beta+\gamma)

Now as \angle BUZ=\angle BPZ and \angle BUZ=90 - \alpha, \angle BPZ = 90 - \alpha.
Let \angle CPZ= \mu. Now \angle CPZ+\angle TPC=\angle BUZ \Rightarrow \mu + 90-(\alpha +\beta)=90-\alpha \Rightarrow \mu = \beta \Rightarrow \angle CPZ = \beta \Rightarrow CYZ = \beta.

As \angle UYP+\angle PYZ = \beta +90 + 90-\beta = 180, U, \ Y and Z must be colinear.

QED

Thanks for any feedback.

 

[disregardthat]

Just bumping this... Hoping someone will take a look. I really appreciate it.

 

[rocomath]

Where is \angle APX or rather point X.

 

[disregardthat]

I thought I had switched all the X's with U's as my program wouldn't accept X as a vertex. It's really U. It's changed now.

 

[rocomath]

What program are you using?

 

[disregardthat]

Geogebra. It's great. It's free and our textbooks webpage has a link to it.

If you want it..: here is a direct link to the setup file.
http://www.geogebra.org/download/InstData/Windows/NoVM/GeoGebra_3_0_0_0_Release_Candidate_1.exe

 

[rocomath]

Woah! This program is cool ... thanks! I wish I knew more geometry to help you, lol. I'm actually self-studying it at the moment, but you're further ahead of me in the game so :( I'm sure someone else will respond in a few!

 

[disregardthat]

It's supposed to be Simson's line, and not Simons line.

 

[naveeniitkgp]

PZ is perpendicular to BC, then angle(BZP) is naturally 90 degree, no need for any assumption!