Start with a triangle

Here's a triangle. You can change it by dragging any of the vertices. Try it now! (If you drag a vertex outside the panel and can't get it back, just refresh the page.)

If you have JavaScript disabled, you should see some static figures below. Our explanations will still work with those.

Two lines and their intersection point

Now we draw two lines. One is the angle bisector at one of the vertices (doesn't matter which one). The other is the perpendicular bisector of the side opposite said vertex.

As you can verify by altering the triangle, two lines coincide if the two sides (not the bisected one) have equal length. Otherwise they will intersect at some point (say P). As we can also verify, P will lie outside the triangle. Verification is not mathematical proof, of course. But proof will follow.

A few more lines

From the intersection point P we now drop perpendiculars to the remaining sides of the triangle. If a side is too short to reach the relevant perpendicular, we just extend the side as needed.

Then we join the point P to the remaining two vertices of the original triangle. Our construction is now complete.

A little detail, which you can ignore: we have used two different line styles for our constructions. All solid lines simply join two points that are already known. For the dotted lines you have to solve for a point. (Before computer graphics, the former kind needed just a ruler, the latter kind needed a ruler and a compass.)

Some congruent triangle pairs

With all the perpendiculars we have drawn, there are several right-angled triangles in our construction. All of them have the intersection point P as one of their vertices. Let us consider the triangles in pairs (shown in turn in green).

First, there are the triangles involving our original angle bisector and the perpendiculars we dropped from P. They have the bisector line itself in common, the bisected half-angles are equal, and each also has a right angle. So these two triangles are congruent. They are also mirror-images of each other, sharing one side.

Then, there are the two triangles involving the perpendicular bisector we drew. They have the bisector itself as a common side, and the bisected sides are equal, and both are right-angled triangles. So again these two triangles are mirror-image congruent triangles.

Now consider the intersection and gap formed when we switch between the above two triangle pairs. These are yet two more right-angled triangles, because each includes one of the perpendiculars we drew. They also share two congruent sides with the above two congruent pairs. Ergo these two triangles are congruent. These are not mirror images, but rotated versions of each other.

One of these green triangles is partly inside the original triangle, one is completely outside. Together they show the difference between the original triangle and an isosceles triangle. You can make make the original triangle isosceles by transferring some length from one leg to the other, and the triangles show you how much.

A sleight of hand

As we have seen, the intersection point P of the angle-bisector line and the perpendicular bisector lies outside the triangle. You can make P appear to lie inside the triangle, by drawing a fake angle bisector. (You could also fake the perpendicular bisector, but that's easier to spot.)

With a genuine perpendicular bisector, the first pair of congruent triangles is still congruent.

The other two pairs of triangles, though they're still right-angled, are not congruent any more. (Unless you make two sides equal.) The fake angle bisector we used to put P inside the triangle breaks the argument. . . 

. . .  which is just as well, because otherwise we'd have proved all triangles to be isosceles!