# How to prove that point E is the intersection of the diagonals of the trapezoid

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.

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

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

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.

$$AD=BC$$

$$AE=EB$$

$$ED=EC$$

$$AE+EC=EB+ED$$

$$AC=BD$$

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.

• I like your approach and your motto, but I've never loved Plato... :D – Cheesecake Apr 3 at 15:36
• Thank you. The Plato thing though was a nickname I got from school. Would you mind NeoPlato? – Nεo Pλατo Apr 3 at 15:38
• No, I wouldn't. NeoPlato won't be the cause of my yikes grades in philosophy. (: – Cheesecake Apr 3 at 15:39
• 😅 I'll take that. – Nεo Pλατo Apr 3 at 15:40
• Once I figure out how to change my name – 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 $$\left(0,\frac{bh}{b+a}\right)$$