While looking at the bisection of area formula (see previous posts)
a’b’ = (1/2)ab
where a’ and b’ are the distances from the vertex of points on the sides of the triangle, and a and b are the lengths of the sides I remembered another formula about triangles, the bisection of the angle formula, with a” and b” being the lengths of the two parts into which the opposite side is divided, namely
a”/b” = a/b
These are like Cuba and Puerto Rico, “Two wings of the same bird”, Jose Marti (in Spanish)
Neither involves the angle itself, and so is very general. I decided that there must be a connection, and after a futile look for some duality in the situation I suddenly saw the connection, in simple algebraic terms:
a’b’ = a’/(1/b’)
and so a triangle with sides a’ and 1/b’ will give a”/b” = a’b’
and then a”/b” = (1/2)ab as well.
This is the construction for the halving a triangle.
This is the extended construction for the bisector. The opposite side is in brown.
and this is a gif showing how as the point a’ (E) is moved the brown line (OE, opposite side) moves parallel to itself, thus preserving the value of the ratio a”/b”
And in case you got this far, some light relief. Bullfrog eats dog food, this morning. The dish is 10 inches across.
I have been developing this computer software / program / application for some years now, and it is now accessible as a web page, to run in your browser.
It provides basic geometric construction facilities, with lines, points and circles, from which endless possibilities follow.
Just try it out, it’s free.
Click on this or copy and paste for later : www.mathcomesalive.com/geostruct/geostructforbrowser1.html
.Here are some of the basic features, and examples of more advanced constructions, almost all based on straightedge and compass, from “make line pass through a point” to “intersection of two circles”, and dynamic constructions with rolling and rotating circles.
Two lines, with points placed on them
Three random lines with two points of intersection generated
Five free points, three generated circles and a center point
Three free points, connected as point pairs, medians generated
Two free circles and three free points, point pairs and centers generated
GIF showing points of intersection of a line with a circle
Construction for locus of hypocycloid
GIF showing a dilation (stretch) in the horizontal direction
Piston and flywheel
Construction for circle touching two circles
Construction for the locus of a parabola, focus-directrix definition.
To show that the angle bisector of an angle in a triangle splits the opposite side in the ratio of the two adjacent sides.
My first proof used angles in the same segment. See
Several tries later (today) and i came up with this annoyingly simple proof:
The theorem states that if we bisect an angle of a triangle then the two parts of the side cut by the bisector are in the same ratio as the two other sides of the triangle. So I thought, what if we move the point, G in the gif below, so that GB always bisects the angle AGC, then where does it go ? Or, more mathematically, what is its locus. Geometry failed me at this point so I did a coordinate geometry version of the ratios and found that the locus was a circle ( a little surprised !). The algebra is fairly simple.
What was a real surprise was that if we take the origin of measurements to be the other point where the circle cross the line ABC, then point C is at the harmonic mean of points A and B. There is YOUR problem from this post.
Getting the construction to show all this was tricky. It is shown below.
I did the construction on my geometry program, Geostruct (see my Software page).
Line bb passes through A. Point D is on line bb. Point E is on this line and also on the line through C parallel to line cc joining B and D, So AD and DE are in the same ratio as AC and CB. When point D is moved the ratios are unchanged (feature of the program). So the point of interest is the intersection of a circle centre A, radius AD, and a circle centre C and radius DE (= FC)
While working on my software for 3D spline curves I needed to find a point between two others which was not halfway.
This turned into finding the ratio which an angle bisector of a triangle splits the opposite side. I worked it out with coordinate geometry and vectors, messy, messy, and then found out that this was a regular theorem in geometry. Here is the geometrical proof which I came up with. It sure ties a few things together: