Monotonicity of a length in a figure Here is something I have been wondering about today... It seems graphically obvious, but I haven't been able to find a "clean" proof of it.

If $x_1>x_2$, then $d_1>d_2$ in the following figure:


Would someone have a clue ?
 A: As $BC=BD$ the perpendicular bisector $m$ of $CD$ goes through $B$. The assumption $x_1>x_2$ means that the point $A$ is to the right of $m$, and the same is then true for the point $E$. It follows that $d_1>d_2$.
A: Applying the cosine Law in $BCE$ and $BDE$ yields an inequality between $\cos (CBE)$ and $\cos(DBE)$. 
Now apply again cosine Law in $BCA$ and $BDA$, again with respect to the angles $CBE$ and $DBE$. 
Edit Here are all details:
$$x_1^2=s^2+BA^2-2sBA \cos(CBE) \,.$$
$$x_2^2=s^2+BA^2-2sBA \cos(DBE) \,.$$
$$x_1 \geq x_2 \Rightarrow \cos(DBE) > \cos(CBE)$$
Now
$$d_1^2=s^2+BE^2-2sBE \cos(CBE) \,.$$
$$d_2^2=s^2+BE^2-2sBE \cos(DBE) \,.$$
Since $\cos(DBE) > \cos(CBE)$ we get $d_1 >d_2$.
A: I think this works:
Let $D(e)=d_1-d_2$, where $0\le e\le 1$ is the proportion $EA/BA$. Clearly (cough), $D$ is continuous, $D(1)=0$, and $ D(0)>0$.  
Now observe that if $D(y)=0$ with $y\ne1$, the triangles $BCE$ and $BED$ would be congruent; and, hence, $D$ would be identically 0.
These observations and the intermediate value theorem imply $D\ge0$ always.
