¶An other way to do the same after Pelitarius, by parallel lines.
Suppose that the circle ge∣uen
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Suppose that the circle ge∣uen
line GH parallel to the side AB, and touching the circle ABC (which is done by a Pro∣position added by Pelitarius after the 17. of the third). And so likewyse draw these lines HK, KL, and LM, parallel to ech of these sides BC, CD, and DE, and touching the circle. And for asmuch as the lines FG and FH fall vpon the two parallel lines AB and GH,* 1.2 the two angles FGH, & FHG, are equall to the two angles FAB and FBA, the one to the other (by the 29. of the first). Wherefore (by the sixt of the same) the two lines FG and FH are equall. And by the same reason, the two angles FHK & FKH, are equall to the two angles FGH and FHG the one to the other: and the line FK is equall to the line FH, and therefore is equall to the line FG. And forasmuch as the angles at the poynt F are equall, therefore (by the 4. of the first) the base HK is equall to the base GH. In like sort may we proue, that the three lines FK, FL, and FM, are equall to the two lines FG and FH. And also that the two bases KL and LM, are e∣quall to the two bases GH and HK: and that the angles which they make with the lines FK, FL, and FM, are equall the one to the other. Now then draw the fift line MG: which shall be equall to
An other way to do the sam•• after Pelita∣rius.
Demonstra∣tion.