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I would like a proof to a Theorem (via OnlineMathLearning.com) I found:

A quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent.

And more specifically, Wikipedia's "Isosceles trapezoid" entry says:

(An isosceles trapezoid is a) trapezoid in which both legs and both base angles are of equal measure.

If necessary, assume that the diagonals bisect the base angles.

I can bring more sources that simply state the properties of an isosceles trapezoid that corroborate the above claims, but no proofs.

I am really struggling finding any proof for the definition of an isosceles trapezoid (and especially for a proof that only starts with the assumption that the diagonals of the quadrilateral are congruent and/or that they bisect the base angles).

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  • $\begingroup$ What are the base angles? A quadrilateral (even one with congruent diagonals) does not have a preferred base. $\endgroup$ Aug 6 at 2:35
  • $\begingroup$ I suspect that the source intended to state that "A trapezoid is an isosceles trapezoid if and only if the diagonals are congruent" (not "A quadrilateral is ..."). Likewise for the other theorems listed in that image. Otherwise, the statements simply aren't true. (Note that Theorem $1$ uses the term "legs", which doesn't apply to general quadrilaterals.) $\endgroup$
    – Blue
    Aug 10 at 20:30

2 Answers 2

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Label the trapezoid as the shown below. The base of the quadrilateral is $\overline{AD}$. The lengths of the sides $\overline{AB},\overline{BC},\overline{CD},\overline{AD}$ are respectively $b,c,d$, and $a$. The diagonals $\overline{AC}$ and $\overline{BD}$ both have length $r$. Angles $\angle BAC$ and $\angle CAD$ are both $\alpha$, while $\angle ADB$ and $\angle BDC$ are both $\beta$.

diagram

Now because of how we define the "inside" of the quadrilateral, it must be the case that $2\alpha<\pi$ and $2\beta<\pi$. Furthermore, since $\overline{AB}$ and $\overline{BC}$ must intersect, $2\alpha+\beta<\pi$. Likewise, $\alpha+2\beta<\pi$.

This valid region for $\alpha$ and $\beta$ is shown below, excluding the boundary.

enter image description here

Note that because of this range of allowable values of $\alpha$ and $\beta$, the sines and cosines of these angles are positive.

Now, by equating expressions for the positions of points $B$ and $C$, we get the following equations.

$$\begin{align*} r\cos\alpha&=a-d\cos2\beta & &(1)\\ b\cos2\alpha&=a-r\cos\beta & &(2)\\ r\sin\alpha&=d\sin2\beta & &(3)\\ a\sin2\alpha&=r\sin\beta & &(4)\\ \end{align*}$$

Rearranging $(1)$ $$a-r\cos\alpha=d\cos2\beta$$

Dividing by $(3)$ $$\frac{a-r\cos\alpha}{r\sin\alpha}=\frac{d\cos2\beta}{d\sin2\beta}$$

Solving for $r$ $$a-r\cos\alpha=r\sin\alpha\cot2\beta$$

$$r=\frac{a}{\cos\alpha+\sin\alpha\cot2\beta}$$

Solving $(2)$ and $(4)$ for $r$ $$r=\frac{a}{\cos\beta+\sin\beta\cot2\alpha}$$

Combining these equations

$$\cos\alpha+\sin\alpha\cot2\beta=\cos\beta+\sin\beta\cot2\alpha$$

Multiplying by $\sin2\alpha\sin2\beta$

$$\sin2\alpha\sin2\beta\cos\alpha+\sin2\alpha\cos2\beta\sin\alpha\\=\sin2\alpha\sin2\beta\cos\beta+\cos2\alpha\sin2\beta\sin\beta$$

Applying double angle theorems $$4\sin\alpha\cos^2\alpha\sin\beta\cos\beta+(1-2\sin^2\beta)2\sin^2\alpha\cos\alpha\\=4\sin\alpha\cos\alpha\sin\beta\cos^2\beta+(1-2\sin^2\alpha)2\sin^2\beta\cos\beta$$

Distributing and factoring

$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(1-\cos^2\alpha)\cos\alpha-2(1-\cos^2\beta)\cos\beta=0$$

$$4\sin\alpha\cos\alpha\sin\beta\cos\beta(\cos\alpha-\cos\beta)-4\sin^2\alpha\sin^2\beta(\cos\alpha-\cos\beta)\\+2(\cos\alpha-\cos\beta)-2(\cos^3\alpha-\cos^3\beta)=0$$

$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2\\-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))(\cos\alpha-\cos\beta)=0$$

Now

$$(4\sin\alpha\cos\alpha\sin\beta\cos\beta-4\sin^2\alpha\sin^2\beta+2-2(\cos^2\alpha+\cos\alpha\cos\beta+\cos^2\beta))<0$$

for all $\alpha$ and $\beta$ in the legal range†. (It is exactly zero at $\alpha=0,\beta=\frac{\pi}{2}$ and vice versa.) This means that

$$\cos\alpha-\cos\beta=0$$ $$\Rightarrow \alpha=\beta$$

Now clearly the base angles $2\alpha$ and $2\beta$ are equal. By SAS congruence, we get that $\triangle ABD$ and $\triangle DCA$ are congruent. Therefore the base side lengths $b$ and $d$ are equal.


Verified graphically. A full proof would likely include function maximization to show that the expression does not exceed $0$.

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  • $\begingroup$ I am absolutely blown away by your proof, and am tremendously grateful! Thank you so much! $\endgroup$
    – Sam
    Aug 11 at 16:18
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Old Answer: Question was asked without angle bisection constraint.

The diagonals of a quadrilateral being congruent is not sufficient to prove that the quadrilateral is an isosceles trapezoid. Consider the quadrilateral shown below:

enter image description here

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  • $\begingroup$ Sorry, left out a detail: If neccesary, assume that the diagonals bisect the base angles. Edited. $\endgroup$
    – Sam
    Aug 5 at 20:55
  • $\begingroup$ Also, according to the source I just added (where I was coming from in asking the question), "a quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent". $\endgroup$
    – Sam
    Aug 8 at 4:11

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