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

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

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Is $\Delta ABC$ equilateral? –  Robert Mastragostino Nov 1 '12 at 22:33
We don't know if it is equilateral. –  Samuel Anthony Nov 1 '12 at 22:59
So maybe if we can't proove that a=b, can we proove that |CB|+|CA|=2|AB|? If yes, how? –  Samuel Anthony Nov 1 '12 at 23:06
If someone had find topic with this question it would be really nice to post a link here. –  Samuel Anthony Nov 1 '12 at 23:37
Does anybody has a link for the topic with this question? I've heard it was discussed on this forum, but I can not find it here. –  user47959 Nov 2 '12 at 17:40

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

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or at least there was not equality in terms of the original question. But there is in the new version of the question –  Henry Nov 2 '12 at 1:45
Could you tell me which program did you use to draw it, please? –  Samuel Anthony Nov 2 '12 at 22:47
@Samuel Anthony: GeoGebra though it took some manipulation to get two points to coincide –  Henry Nov 2 '12 at 23:17

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: $$2|AB| =2a+2b = (a+x)+(b+y)=|CA|+|CB|.$$

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