Geometry - Simon's line

1. Jan 26, 2008

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.

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• Simson line.JPG
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Last edited: Jan 27, 2008
2. Jan 27, 2008

disregardthat

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

3. Jan 27, 2008

rocomath

Where is $$\angle APX$$ or rather point X.

4. Jan 27, 2008

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.

5. Jan 27, 2008

rocomath

What program are you using?

6. Jan 27, 2008

disregardthat

7. Jan 27, 2008

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!

8. Jan 28, 2008

disregardthat

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

9. Mar 13, 2011

naveeniitkgp

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