# 3 Isosceles Trapezoids

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A Mathematical Droodle

Let there be 6 points O_{1}, O_{2}, O_{3}, O_{4}, O_{5} and O_{6}. Assume also that in the hexagon O_{1}O_{2}O_{3}O_{4}O_{5}O_{6} there are three pairs of parallel sides: _{1}O_{2} || O_{4}O_{5},_{2}O_{3} || O_{5}O_{6}_{3}O_{4} || O_{6}O_{1}

(Due to this statement, it is impossible to have all three trapezoids isosceles unless all three circles coincide. The applet, therefore, insures that the red and the blue quadrilaterals are isosceles trapezoids. The green quadrilateral acquires that symmetry if and only if the three circles coincide.)

### Lemma

Let each side of the hexagon O_{1}O_{2}O_{3}O_{4}O_{5}O_{6} be parallel to its opposite mate. Then the hexagon is inscribable in a circle if and only if the three pairs of opposite sides define isosceles trapezoids.

### Proof ^{(1)}

First, assume that the three quadrilaterals O_{1}O_{2}O_{4}O_{5}, O_{2}O_{3}O_{5}O_{6} and O_{3}O_{4}O_{6}O_{1} are all isosceles trapezoids. The circumcenter of the first lies on the common perpendicular bisector of O_{1}O_{2} and O_{4}O_{5}. The circumcenter of O_{2}O_{3}O_{5}O_{6} lies on the common perpendicular bisector of O_{2}O_{3} and O_{5}O_{6}. Consider the point S of intersection of the two lines. I wish to show that all six points are equidistant from S.

By construction, points O_{1} and O_{2} certainly are (from the first trapezoid) as are points O_{2} and O_{3} (from the second trapezoid). Also by construction, _{4} = SO_{5}_{5} = SO_{6}

In the third trapezoid, O_{3}O_{4}O_{6}O_{1}, we therefore have

SO_{1} = SO_{3} and SO_{4} = SO_{6}

so that the perpendicular bisectors of the sides O_{1}O_{3} and O_{4}O_{6} cross at S, and, because of the symmetry of the isosceles trapezoid O_{3}O_{4}O_{6}O_{1}, the common perpendicular bisector of its bases, O_{3}O_{4} and O_{6}O_{1}, also passes through S.

This completes the "if" portion of the proof. The "only if" portion is obvious. If a hexagon inscribed in a circle has the sides parallel to their opposites, then the opposite sides do of course define isosceles trapezoids.

### Proof ^{(2)}

The sides of an isosceles trapezoid are equal and so are its diagonals. Any two of the three trapezoids at hand share a diagonal, such that in all there are just three segments: O_{1}O_{4}, O_{2}O_{5}, and O_{3}O_{6}. Any two of the segments serve as diagonals of one of the three trapezoids and are, therefore, equal. Which implies that all three are equal. Having established that, we are now in a position to apply Problem 109 from [Shklyarsky]:

Prove that, if in a hexagon the opposite sides are parallel and the three diagonals joining the opposite vertices are equal, then the hexagon is inscribable in a circle.

A trapezoid with equal diagonals is necessarily isosceles. The common perpendicular bisector of O_{1}O_{2} and O_{4}O_{5} serves as the axis of symmetry of the trapezoid O_{1}O_{2}O_{4}O_{5}. It passes through the point of intersection of the diagonals O_{1}O_{4} (or their extensions) and O_{2}O_{5} and bisects the angle formed by the two. Similar conclusions could be drawn regarding the other two trapezoids. The three diagonals form a triangle in which the axes of symmetry of the three trapezoids bisect internal angles. The axes of symmetry are concurrent at the incenter of the triangle formed by the three diagonals. Naturally, all six points O_{1}, ..., O_{6} are equidistant from that point.

### References

- Floor van Lamoen,
*Private communication*, May 2002. - D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom,
*Selected Problems and Theorems of Elementary Mathematics*, v 2, Moscow, 1952

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