Then, since RO is at right angles to the plane of the circle LMN,
therefore RO is also at right angles to each of the straight lines LO, MO, NO.
And, since LO is equal to OM, while OR is common and at right angles,
therefore the base RL is equal to the base RM. [I. 4]
For the same reason RN is also equal to each of the straight lines RL, RM;
therefore the three straight lines RL, RM, RN are equal to one another.
Next, since by hypothesis the square on OR is equal to that area by which the square on AB is greater than the square on LO,
therefore the square on AB is equal to the squares on LO, OR.
But the square on LR is equal to the squares on LO, OR,
for the angle LOR is right; [I. 47]
therefore the square on AB is equal to the square on RL;
therefore AB is equal to RL.
But each of the straight lines BC, DE, EF, GH, HK is equal to AB, while each of the straight lines RM, RN is equal to RL;
therefore each of the straight lines AB, BC, DE, EF, GH, HK is equal to each of the straight lines RL, RM, RN.
And, since the two sides LR, RM are equal to the two sides AB, BC, and the base LM is by hypothesis equal to the base AC,
therefore the angle LRM is equal to the angle ABC. [I. 8]
For the same reason the angle MRN is also equal to the angle DEF, and the angle LRN to the angle GHK.