Imagine a triangle.

Let’s divide the angle at point *A* in half.

The dotted line is called the *bisector*. The bisector divides the opposite line segment in a part of length *x* and a part of length *y.*

#### Equal ratios

The Angle Bisector Theorem states that the ratio *x : b* equals the ratio *y : c*. Or equivalently, *b/x = c/y.* This means that if *b *is for example twice the length of *x*, then *c* is twice the length of *y.*

In the drawing below, measuring in millimeters, I found *b/x* = 32/25 (= 1.28), while *c/y* = 61/48, which is roughly1.27*. *So my drawing is not so bad in that respect.

#### Who needs proof?

Several proofs of this theorem exist. My interest however is to find a construction that makes immediately clear that the theorem holds. I just want to *see* that it is true. And from there, effortlessly I hope, write down the proof if anyone needs it.

#### First attempt: Pairs of congruent triangles

Use compass and ruler (if you like) to construct points *P, Q* and *R*.

There are two pairs of congruent triangles: *△ABR* (the large triangle containing all geometry) to *△APC* (the little one crouching at the top) and *△PCQ* to *△RBQ* (the green ones). From these it follows that *x : y = PC : BR = b : c*.

So *x : b = y : c.*

It’s not complicated, but it requires too many steps. I would like something simpler.

#### Second attempt: One congruency less

Another possibility is to extend *△ABD* with the isosceles triangle *△BDE* as above. The two coloured triangles have two angles in common (indicated with open and closed discs) and so are congruent.

So now we have congruent triangles with *x* and *b* in one triangle and *y* and *c* in the other. Now it’s obvious that *x : b = y : c.*

#### Third attempt: Simpler congruency

Here’s another, even simpler one, arriving at two triangles with two equal angles, therefor congruent.

We got rid of the reflection. The congruency consists of a scaling and a rotation operation only.

#### Final attempt

In the previous attempt it appeared the reflection part of the congruency wasn’t needed. Well, the rotation isn’t needed either:

No rotation or reflection, just a scaled (and perhaps translated – depends on the transformation origin) copy.

#### And the winner is…

I’m satisfied with that, but if you find alternatives, please let me know! Actually, I think my preference goes to attempt number 3.