Draw a regular nonagon. If we label the vertices of the polygon sequentially, starting from A, we can draw three line segments: AB, AC, and AE.

Show that AB + AC = AE.

A mildly confusing geometric demonstration…

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Make 4 identical nonagons, and Label the vertices in each from A to I.  I have the horizontal bases all as GH in my orientation.  Line the nonagons up so that G in the 2^nd one is coincident with C in the 1^st, H in the 3^rd is coincident with F in the 2^nd, and B in the 4^th is coincident with G in the 3^rd.  This will make I in the 4^th coincident with E in the 1^st one.

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For ease of description let A in the 1^st be P, G in the 2^nd/H in the 3^rd be Q, D in the 3^rd be R and I in the 4^th/E in the 1^st be S.

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We can then make the quadrilateral PQRS.  The lengths PQ and QR are both made up of an AC equivalent diagonal and a side length (AB equivalent), and hence are equal, and RS and PS are both made of an AE equivalent length. Hence PQRS is a kite.

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Further, it is easy to go through and show that both angle PQR, and angle PSR are both equal to 140 degrees.  If the angles between the pairs of equal length sides are equal, then the kite is a rhombus, and PQ = QR = RS = PS.

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PQ was the length of AC + AB, and PS was the length of AE so AB + AC = AE.

 

Pardon my clumsiness at putting a simple picture into words, but draw it and you'll see it's obvious...

Posted by Alec on 2004-11-26 15:17:53