Re: How to compute triangle base/altitude intersection

Pascal Bourguignon said:

"Arthur J. O'Dwyer" <ajonospam@xxxxxxxxxxxxxx> writes:

On Wed, 30 Aug 2006, Pascal Bourguignon wrote:
"Phlip" <phlipcpp@xxxxxxxxx> writes:
Balabek wrote:

Hope you can help me solve this problem.

Suppose I have a triange ABC. I have altitude passing through vertice
C, and it intersects line AB or extension of line AB at point D.

How do you define "altitude"?

An altitude of a triangle passes through one of the vertices and
is perpendicular to the opposite side. I don't know what the word for
"altitude" is in French, but I would expect it would be a cognate to
the English word "altitude".

Oh, it's called "la hauteur du triangle" (height).
"Altitude" in French is usually absolute, not relative.

Ok, so you have the normal problem, not a strange variation.

There's no difficulty in it. As have been explained, to find the
position of C geometricaly you just draw two circles centered on A and
B of the given radius, and find their intersections. When you draw
the line CC', the intersection of this line with AB is D.

No, I thought that too, but it's wrong, or rather it's only true if ABC is
isosceles, with AC and BC being equal in length. Consider this
counter-example:

C1
+
+ + +
+ + +
A + D+ + B
+++++++++++++++++++++++++++++++++
+ + +
+ + +
+ + +
+
C2

Clearly neither C1D nor C2D is perpendicular to AB. What we really want is
this:


C1
+
+ | +
+ | +
A + D1 | + B
+++++++++++++++++++++++++++++++++
+ | D2 +
+ | +
+ | +
+
C2

I have explained how to achieve it elsethread.

--
Richard Heathfield
"Usenet is a strange place" - dmr 29/7/1999
http://www.cpax.org.uk
email: rjh at above domain (but drop the www, obviously)
.

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