# 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 ?

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This is a job for the Hinge Theorem! (proofwiki.org/wiki/Hinge_Theorem) Apply the theorem's converse to $\triangle ABC$ and $\triangle ABD$ to get that $\angle ABC > \angle ABD$; then apply the theorem "forwards" to $\triangle EBC$ and $\triangle EBD$. So, the obvious question: How do you prove the Hinge Theorem? The answers you get here should do it, and there's the ProofWiki link, but now you also have a name you can Google. –  Blue Nov 22 '11 at 12:17

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 $d_1>d_2$.

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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 $d_1 >d_2$.

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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.

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