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

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

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

Let [tex]\angle BZP=\alpha[/tex] and [tex]\angle APU=\beta[/tex].

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

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

Let [tex]\angle UZB = \gamma \Rightarrow \angle UPB = \gamma[/tex].

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

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

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

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

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

[tex]QED[/tex]

Thanks for any feedback.

 

[disregardthat]

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

 

[rocomath]

Where is [tex]\angle APX[/tex] 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!