of DE, deſcribe the circle DEH; and let DA ee 2. poſt. be produced to the point G in the circumference thereof. Then AG = CB.

For DG ff 15. def.
g conſtr.
h 2. ax.
k 15. def.
l 1. ax. = DE, and DA g = DC. Wherefore AG b = CE k = BC l = AG. Which was to be done.

The putting of the point A within or without the line BC varies the caſes; but the conſtruction, and the demonſtration, are every where alike.

The line AG might be taken with a pair of compaſſes; but the ſo doing anſwers to no Poſtulate, as Proclus well intimates.

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Two right lines, A and BC, being given, from the greater BC to take away the right line BE equal to the leſſer A.

At the point B aa 2. 1. draw the right line BD = A. The circle deſcribed from the center B at the diſtance of BD ſhall cut off BE bb 15. def.
c conſtr.
d 1. ax. = BD c = A d = BE. Which was to be done.

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If two triangles BAC, EDF, have two ſides of the one BA, AC equal to two ſides of the other ED, DF, each to its correſpondent ſide (that is, BA = ED,