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therefore the rectangle BA, AC is equal to the square on AD. [VI. 17]

And since, as AB is to BC, so is EB to BF, [VI. 1] and EB is the rectangle BA, AC, for EA is equal to AC, and BF is the rectangle AC, CB, therefore, as AB is to BC, so is the rectangle BA, AC to the rectangle AC, CB.

And the rectangle BA, AC is equal to the square on AD, and the rectangle AC, CB to the square on DC, for the perpendicular DC is a mean proportional between the segments AC, CB of the base, because the angle ADB is right. [VI. 8, Por.]

Therefore, as AB is to BC, so is the square on AD to the square on DC. Q. E. D.

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