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Consider the trapezium ABCD in which AD and BC are the top and the bottom - parallel and of known length - and AC and BD are the diagonals - also of known length. Suppose AC and BD intersect at O.

Then, it can be shown that triangles AOD and COB are similar. Therefore AO/OC = DO/OB = AD/BC where both lengths for the last ratio are known. Then, given AC, it is possible to calculate OC (and AO) and given BD, it is possible to calculate OB (and DO).

So all sides of triangle COB are known and so it is easy to construct it. Then simply extend BO to D (adding OD) and CO to A (adding OA). Join BA, AD and DC. Done!

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11y ago

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