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Q. Let $ABC$ be a triangle and P be any point on $(ABC)$. Let X, Y , Z be the feet of the perpendiculars from P onto lines BC, CA, and AB respectively. Prove that points $X, Y , Z$ are collinear.


Actually I started going through the book EGMO and got an alternate proof.But I guess, it's wrong somewhere please check my steps...

Answer-
Extend XY such that it meets AB extended, at Z'
It would be suffice to show that AYPZ' is cyclic.
Also join PC
$∡Z'YP = ∡XYP $
$∡Z'AP=∡BAP=∡XCP$
It's easy to observe that PYXC is also cyclic
$∡XYP=∡XCP$ Hence, $∡Z'AP = ∡Z'YP$ And so, $ Z'=Z$ proved

Sauhardya Hazra
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  • Draw the figure so that $Z$ is on the ray $[BA$, but not on the segment $[BA]$. Then there are some issues with the above relations. In other words, the above rows depend on the chosen figure. Else: Please use mathjax to properly format the math symbols. Link: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference . So use $ABC$ instead of ABC and so on. The angles can be typed in a sample as $\sphericalangle Z'YP$ which looks like $\sphericalangle Z'YP$. And which is the one question? Please state it explicitly. – dan_fulea Apr 05 '22 at 13:48
  • I am extremely sorry for the mess, that spherical angle notation I have used is for directed angles... – Sauhardya Hazra Apr 06 '22 at 14:26
  • This notation is not the problem. (Well it would have been nice to have a posted question using mathjax, but somebody will edit the question soon...) The main problem i have is as follows. Draw a picture so that $P$ is on the arc $\overset\frown{AC}$ and so that the projection $Z$ (which a posteriori is also $Z'$) of $P$ on $AB$ is not inside the segment $AB$, but where $X$ is in the segment $BC$, and $Y$ is in the segment $AC$. Then the relation $$\widehat{Z'YP}=\widehat{XYP}$$is not valid (in general). So the proof depends on the picture. (Inserting a picture is a good idea in such cases.) – dan_fulea Apr 06 '22 at 14:49
  • But I don't get it, then how can such a quadrilateral ABPC be cyclic...Please give a better example – Sauhardya Hazra Apr 07 '22 at 14:29
  • And also, may be there is something wrong with my proof or maybe there's a confusion as I cannot post a diagram, but the statement that you made cannot be correct, I guess...since it is a proof of Simson's line, right – Sauhardya Hazra Apr 07 '22 at 14:35
  • OK, in order to have here a quick finish, please state explicitly a question. (This is a questions and answers site.) When the question is stated and clear, i will answer it, there is much more space in the answers field. – dan_fulea Apr 07 '22 at 14:45
  • I have tried my best to make it as simple and explicit as possible, please check it now... – Sauhardya Hazra Apr 07 '22 at 15:43

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