Proving that $|CA|+|CB|=2|AB|$ in a general $ABC$ triangle

Question

How in this situation (presented in image) can I prove that $|CA|+|CB|=2|AB|$?

Answer 1

Here is an example where there is not equality (distances are rounded)

Answer 2

Here's a calculatory approach:

Call $AE=x$ and $BD=y$ and let the angles at $A$, $B$, $C$ be $\alpha$, $\beta$, $\gamma$, respectively.

Now, the area of the triangle $ABC$ equals $$\frac 12 (a+x) (b+y) \sin \gamma.$$

But it is also the sum of the three smaller regions:

$$ab\sin\gamma +\frac 12 ay\sin\beta +\frac 12 xb \sin \alpha.$$

Now, equate the two expressions, divide everything by $\sin \gamma$ and use $\dfrac{\sin \beta}{\sin \gamma} = \dfrac{a+x}{a+b}$ and $\dfrac{\sin \alpha}{\sin \gamma} = \dfrac{b+y}{a+b}$ to obtain

$$ab+ \frac 12 ay \frac {a+x}{a+b} + \frac 12 xb \frac{y+b}{a+b} = \frac 12 (a+x)(b+y).$$

Multiply everything by $2(a+b)$ and develop to get:

$$a^2b+ab^2=aby+xab.$$

Divide by $ab$ to get

$$a+b=x+y$$

which is equivalent to the required equation: $$2|AB| =2a+2b = (a+x)+(b+y)=|CA|+|CB|.$$