How in this situation (presented in image) can I prove that $|CA|+|CB|=2|AB|$?
Proving that $|CA|+|CB|=2|AB|$ in a general $ABC$ triangle
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0Does 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. – 2012-11-02
2 Answers
Here is an example where there is not equality (distances are rounded)
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0@Samuel Anthony: [GeoGebra](http://www.geogebra.org) though it took some manipulation to get two points to coincide – 2012-11-02
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: $2|AB| =2a+2b = (a+x)+(b+y)=|CA|+|CB|.$