All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 On the same line (Posted on 2013-12-22)
Let ABC be a triangle.
Let the point P lie in the plane of ABC.
Let X, Y, Z be the feet of the perpendiculars from P to the lines BC, AC, AB respectively.

PROVE:  : The points X, Y , Z are collinear if and only if P lies on the circumcircle of ABC.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(2): Solution Comment 3 of 3 |
(In reply to re: Solution by Bractals)

Since the proposition states that Z is the foot of a perpendicular from P
to AB etc., it’s a moot point as to whether we include the special case
of P on AB. But clearly, if we do, and if P coincides with A, then so must
Y and Z, rendering the collinearity of XYZ trivial.

In the ‘converse’, the possibility of XYZ coinciding with a side, say BC,
is more interesting: we would then have Z & B coincident and Y & C
coincident so that ABPC (i.e. AZPY) is clearly cyclic.

You’re right Bractals, I should have included this second point in my
original post.

 Posted by Harry on 2013-12-24 13:18:38

Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information