T H E E L E M E N T S	Book III   Proposition 35
	For the same reason, also,     the rectangle DE, EB together with the square on FE is equal to the square on FB. But     the rectangle AE, EC together with the square on FE was also proved equal to the square on FB; therefore     the rectangle AE, EC together with the square on FE is equal to the rectangle DE, EB together with the square on FE. Let     the square on FE be subtracted from each; therefore     the rectangle contained by AE, EC which remains is equal to the rectangle contained by DE, EB. Being what it was required to prove.
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