Re: How to compute triangle base/altitude intersection

Richard Heathfield wrote:
Pascal Bourguignon said:


<snipped>


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

Maybe I'm missing something, or my math is rusty, but if you
found C1 and C2 using circles as described above,
then :
BC1==BC2 (both are radii of the same circle centered at B).
BD<BC1

That makes line segment C1C2 perpendicular by definition. Your
example above would look like this:



C1
+ | +
+ | +
+ | +
+ | +
+ | +
A + | + B
++++++++++++++++++++++++++++++++++
+ | +
+ | +
+ | +
+ | +
+ | +
+ | +
C2

this:


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


This is impossible, although ...

I have explained how to achieve it elsethread.

.... your explanation is correct.

goose,
Welcome back :-)

.