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An isosceles $ABCD$ trapezoid and a straight line $k$ being its axis of symmetry are given. On the straight line $k$ we choose the point $E$ so that the triangle $ADE$ has the smallest perimeter. How to prove that point $E$ is the intersection of the diagonals of the trapezoid.

Maybe there is a simple way but I dont know how to do it. This is reminiscent of the famous exercise about the shortest path that passes by a river.

  • $\begingroup$ You can use the fact that only an isosceles trapezoid can be a cyclic quadrilateral. $\endgroup$ – Cheesecake Apr 3 at 14:25
  • $\begingroup$ I know this fact but I dont kow how to use it. $\endgroup$ – piteer Apr 3 at 14:27
  • $\begingroup$ How you name the vertices? clockwise or counter clockwise? $AB$ is the longest base or $CD$? $\endgroup$ – Qurultay Apr 3 at 14:36
  • $\begingroup$ Ive added a drawing. $\endgroup$ – piteer Apr 3 at 14:39
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    $\begingroup$ @ms._VerkhovtsevaKatya I see now. Thank you. $\endgroup$ – Allawonder Apr 4 at 19:13

Note that $AE+ED \geq AC$ holds (Think about the reflection with respect to $k$ !).

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Before we begin, let us dip our fingers into an interesting minimization problem.

Given a line segment $AB$ and another line, how do we pick a point $E$ on that line such that $|AE|+|EB|$ is minimized?

The key thing is to reflect $B$ on that line as I have done. Then $BE=B'E$ by the laws of reflection.

Let that point $E$ wander around and contemplate the combination of lines $AE+EB'$. How would we minimize this trip from $A$ to $B'$?

Did someone say a straight line? Nice work!

Check that green line. You really can't get any smaller than that. Hence these steps are ideal for such a problem:

  1. Reflect $B$ on the line (or $A$, it doesn't really matter)
  2. Join the reflected point with $A$ by a straight line.
  3. The intersection of both lines is the desired point.

Using this in the problem at hand:

Wait! Wait just one moment. It says minimum perimeter. But $AD$ is a given unless you want to morph it thus our job is to minimize $AE+DE$

And we just follow our discovered steps.

Reflect $D$ on line $k$. That line is by definition the symmetry line of the trapezium $ABCD$. Therefore $D'=C$

Join $C$ and $A$. This segment intersects $k$ at $E$. Q. E. D!

Well, almost. Let's clear this proof.






And since $AED$ and $CEB$ are hence congruent, $\widehat{AED}=\widehat{BEC}$.

And by this creation of vertically opposite angles, $BD$ is as much a straight line as $AC$.

They meet at $E$ as was already shown and hence thy conjecture.

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  • $\begingroup$ I like your approach and your motto, but I've never loved Plato... :D $\endgroup$ – Cheesecake Apr 3 at 15:36
  • $\begingroup$ Thank you. The Plato thing though was a nickname I got from school. Would you mind NeoPlato? $\endgroup$ – Nεo Pλατo Apr 3 at 15:38
  • $\begingroup$ No, I wouldn't. NeoPlato won't be the cause of my yikes grades in philosophy. (: $\endgroup$ – Cheesecake Apr 3 at 15:39
  • $\begingroup$ 😅 I'll take that. $\endgroup$ – Nεo Pλατo Apr 3 at 15:40
  • $\begingroup$ Once I figure out how to change my name $\endgroup$ – Nεo Pλατo Apr 3 at 15:47

Let's solve it analytically.

Suppose $AB=b$ and $CD=a$ and let the height be $h$. Then we can put the trapezoid in location $A=(-b/2,0)$, $B=(b/2,0)$, $C=(a/2,0)$ and $D=(-a/2,0)$.

$y=0$ is the symmetric axis and therefore $E=(0,y)$.

We need to minimize the expression $$AE+ED=\sqrt{\frac{b^2}{4}+y^2}+\sqrt{\frac{a^2}{4}+(y-h)^2}$$ derivationg with respect to $y$ and simplifying, gives $$(b^2-a^2)y^2-2b^2hy+b^2h^2=0$$ which has two roots $$y=\frac{bh}{a+b}\quad\text{and}\quad y=\frac{bh}{b-a}$$ On the other hand, the intersection of two diagonals is the point $$\left(0,\frac{bh}{b+a}\right)$$

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