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Now the rectangle CF, FA is FK, for AF is equal to FG; and the square on AB is AD;

therefore FK is equal to AD.

Let AK be subtracted from each;

therefore FH which remains is equal to HD.

And HD is the rectangle AB, BH, for AB is equal to BD; and FH is the square on AH;

therefore the rectangle contained by AB, BH is equal to the square on HA. therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB, BH equal to the square on HA. Q. E. F.

Proposition 12.

In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle, namely that on which the

Euclid. Euclid's Elements. Sir Thomas Little Heath. New York. Dover. 1956.

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