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Here's one question that has been bothering me now for a while. It is not homework.

For an isosceles trapezoid (wikipedia link:, do we always have that sum of the two diagonals is larger (or equal) than sum of the two bases? No matter how we choose the legs and the bases?

I tried this for many examples on the plane and it always seemed to be right (if my calculations were correct) and intuitively it seems to be correct. How would one go about proving this claim if its true? And if it isn't, does someone have a counter-example in mind?

Thanks for all the input in advance.

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Did you try to use the triangular inequality? – Sigur Sep 11 '12 at 11:37
@Sigur: Do you mean the triangle inequality? I hadn't so far, but if $a,b$ are bases and $c$ the leg, $d$ the diagonal, then triangle inequality would give $a+b\leq c+d+c+d=2d+2c$. The $2c$ kind of ruins it:( Or did I misunderstand your point? – T. Eskin Sep 11 '12 at 11:42
A hint: think about the point where the two diagonals intersect, and the two triangles defined by that point and the two bases. – Adam Bailey Sep 11 '12 at 11:58
up vote 1 down vote accepted

Let the trapezoid be $ABCD$, where $AB$ is the longer base and $CD$ is the shorter base. Let $E$ and $F$ be the orthogonal projections of $C$ and $D$, respectively, onto the line $AB$. Since your trapezoid is isosceles, these points will lie between $A$ and $B$.

The diagonal $AC$ is longer than $AE$, and the diagonal $BD$ is longer than $BF$. But the sum of $AE$ and $BF$ is equal to the sum of the two bases.

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That's a really nice and clear way to see it. Thanks. – T. Eskin Sep 11 '12 at 12:29
In fact, you can drop the assumption that the trapezoid should be isosceles, but then there will be more cases to consider according to the order of the points, $A,B,E,F$. – Per Manne Sep 11 '12 at 14:08

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