# A brain teaser with an Euclidean origin

It’s time for a brain teaser:

There is a triangle $\triangle ABC$, and $D$ is an arbitrary interior point of this triangle (see Fig. 1). Prove that $AD + DB < AC + CB$. Fig. 1

Here is my solution:

Extend line $AD$ to point $E$ on $CB$ (see Fig. 2), Fig. 2

we have $AD + DB < AD + (DE + EB)\quad\quad\because \triangle DEB: DB < DE + EB$ $= (AD + DE) + EB$ $=AE + EB\quad\quad\because AD + DE = AE$ $< (AC +CE) +EB\quad\quad\because \triangle ACE: AE < AC +CE$ $= AC + (CE + EB)$ $= AC +CB\quad\quad\because CE + EB = CB$ $\implies AD +DB < AC +CB$

My solution relies on a well known theorem:

Given a triangle ABC, the sum of the lengths of any two sides is greater than the length of the third side.

In the words of Euclid: “In any triangle two sides taken together in any manner are greater than the remaining one” (The Elements: Book I: Proposition 20)

I have conjured up the following algebraic proof of Euclid’s proposition:

Any $\triangle ABC$ can be put in a rectangular coordinate system where $x_1 > 0, x_2 \ge 0, y_2 > 0$ (see Fig. 3) Fig. 3

It follows that $AB + AC = \sqrt{(x_2-(-x_1))^2+y_2^2} + \sqrt{(x_2-x_1)^2+y_2^2}$ $= \sqrt{(x_2+x_1)^2+y_2^2} + \sqrt{(x_2-x_1)^2+y_2^2}$ $> \sqrt{(x_2+x_1)^2} + \sqrt{(x_2-x_1)^2}\quad\quad\quad\because y_2 > 0$ $\geq \sqrt{(x_1)^2} + \sqrt{(-x_1)^2}\quad\quad\quad\because x_2 \geq 0$ $= 2 |x_1|$ $= BC$ $\implies AB+AC > BC$