I, Carlo Séquin, will prove to you that any triangle that you draw is equilateral.
That means it has the same length of the sides all around.
So I'm starting with three intersecting lines,
and a second one here,
and a third one here,
and I mark the intersection vertices as the vertices of a triangle:
A, B, and C.
That does not look, you know, equilateral (not too far off), but it is certainly arbitrarily drawn.
Now, I'm taking the middle point of the side BC (I call this M), and I draw the perpendicular bisector.
So I'm drawing a line that's perpendicular, coming out of M, going off in that particular direction.
And, of course, we have a right angle here.
Now you know that this line is the locus of all the points
that are equidistant from point B and from point C.
I'm going to draw another extra line, and that is the angle bisector coming out of A here,
So that should look something like that, I guess.
and this line, of course, let's call it angle divider "d",
is the locus of all the points that are equidistant from lines AB and from line AC.
So,
If I'm going to the intersection point between these two lines, which is about here, which I'm calling X,
then I know that this point is equidistant
from B and from C, because it lies on the perpendicular bisector.
So, let's draw these two lines.
And we know that these are equal because they are on the perpendicular bisector.
Okay,
But also,
If I draw the shortest distance from X to line AB,
Which might look something like that,
And comes in, of course, at a right angle here.
and I draw the shortest distance from X on to line AC, and has a right angle here.
Let's call this point B*,
and let's call this point C*,
then I know that XB* is equal to XC*
because they're on the angle bisector,
and these are all the points that are equidistant from those two lines.
Okay.
Now, we focus on these two small triangles.
And here, we have a triangle with a right angle;
adjacent to the right angle we have this double-marked line, XB*.
And here we have a right angle;
and adjacent to the right angle we have this double-marked line, XC*;
And we know these two lines are the same length.
We also have the third side of a triangle which is opposite of the right angle,
marked with this one slash, and it is the same as this slash over here,
(that's the hypotenuse in this case)
And so, because two triangles have two equal sides,
and they have the longer side being opposite to an angle that also happens to be equal,
then these two triangles are congruent.
As a matter of fact, in this case, we can see they're basically mirror images of one another.
Now, this of course tells us that,
we also now know that,
BB*
BB* must be equal to
BB* must be equal to CC*.
BB*
(I'll mark it with three slashes)
BB* is equal to CC*.
Now, we have more congruent triangles.
We also have the triangle
We also have the triangle A
We also have the triangle AX
We also have the triangle AXB*
We also have the triangle AXB*, being a mirror image of
We also have the triangle AXB*, being a mirror image of AXC*
And so those two triangles are congruent.
And this, of course, tells us that
AB* equals to AC*
Okay! Now, what can we conclude from all that?
Well, we can say that
AB* minus BB* is really just the side AB.
And we can say that
AC* minus CC* is just the side AC.
Now we're almost there.
We have proven up here that AB* is equal to AC*,
we have proven that BB* is equal to CC*,
and so these two sides being equal,
it essentially then follows that AB equals AC.
So, we have done nothing special about the triangle
and we have now proven that these two sides are of equal length.
(From the background) So we've got an isosceles triangle?
That's right.
And now or course, I can go around and do exactly the same construction
on two other sides, making this what was A before,
and proving that these two sides are the same
with the same construction hanging off to this side.
I don't have to do it a third time, because
once I've proven that, I can really show that all sides of this triangle are of equal length.
I can just start from another vertex,
draw my angle bisector,
do my perpendicular bisector over there and
do exactly the same construction.
So, you of course know that this cannot possibly be true,
and so I encourage you to go very carefully,
step-by-step, through here.
and try to find out, you know,
where is that one bogus step?
So, be vigilant! Go for it!