# Isosceles trapezoid

I was solving an exercise on Isosceles trapezoid whose diagonal was given, and I note that If I draw a diagonal in the isosceles trapezoid I got two triangles

To determine the area of the triangles I draw their heights, which are perpendicular to the diagonal. The problem arises when I suppose that the sum of the height of the two triangles is equal to the diagonal. Funny thing I got the correct anwser doing that way, however it was only intuitive, not accurate.

But how to prove if $(a = b + c)$ is true or false ?

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Your claim: $a=b+c$ is not true. Since your trapezoid is isosceles, the two lower angles are the same. Therefore, the two diagonals are equal because the lower triangle each creates when dividing the trapezoid are congruent (by SAS). Then, by the Pythagorean theorem, as long as the two perpendiculars are not collinear, $a>b+c$:

Only when the two perpendiculars are collinear will $a=b+c$, but then your isosceles trapezoid will be a square. (As noted by Issac, the two perpendiculars can be collinear with out the isosceles trapezoid being a square.)

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It is not true. The area of the trapezoid is $a(b+c)/2$. Take the trapezoid $(-2,0), (2,0), (1,1), (-1,1)$. This has area $3$ and diagonal $\sqrt{10}$. If $a=b+c$, the area would be $\frac{a^2}{2}=5$ instead of $3$

An easy way to see it is to imagine a very wide trapezoid. $b$ and $c$ will be almost vertical and no longer than the height, but the diagonal $a$ is very long.

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That $a=b+c$ is not true in general is well-covered in other answers. There are, however, numerous isosceles trapezoids for which it is true.

In any isosceles trapezoid $TRAP$ as shown with $TP=RA$, $\triangle TPD\cong\triangle RAD$ and $\triangle RAD$ is the reflection image of $\triangle TPD$ over the line through $D$ perpendicular to $\overline{TR}$ and $\overline{AP}$.

Conversely, starting with any $\triangle XYZ$ and reflecting it over the line through $Z$ for which the acute angles formed by this line and $\overline{XZ}$ and $\overline{YZ}$ have equal measure to $\triangle X'Y'Z$ yields an isosceles trapezoid $XX'Y'Y$ with $Z$ at the intersection of the diagonals.

(‡ This restriction on the angles formed by the line and the sides of the triangle ensures that $X$, $Z$, and $Y'$ are collinear and $X'$, $Z$, and $Y$ are collinear, so that $Z$ is the intersection of the diagonals.)

Now, if you start with any $\triangle XYZ$ with a right angle at $Z$, you will get an isosceles trapezoid with perpendicular diagonals.

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