other angles to the other angles to which the equal sides
are opposite ; [I. 4.
therefore the angle CBF is equal to the angle CDF.
And because the angle CDE is double of the angle
CDF, and that the angle CDE is equal to the angle CBA,
and the angle CDF is equal to the angle CBF,
therefore the angle CBA is double of the angle CBF;
therefore the angle. ABF is equal to the angle CBF;
therefore the angle ABC is bisected by the straight line BF.
In the same manner it may be shewn that the angles
BAE, AED are bisected by the straight lines AF, EF.
From the point F draw FG, FH, FK, FL, FM perpen-
diculars to the straight lines AB, BC, CD, DE, EA. [1. 12.
Then, because the angle FCH is equal to the angle
FCK,
and the right angle FHC equal to the right angle FKC;
therefore in the two triangles FHC, FKC, there are two
triangles of the one equal to two angles of the other, each to
each;
and the side FC, which is opposite to one of the equal
angles in each, is common to both ;
therefore their other sides are equal, each to each, and
therefore the perpendicular FH is equal to the perpen-
dicular FK. [I. 26.
In the same manner it may be shewn that FL, FM, FG
are each of them equal to FH or FK.
Therefore the five straight lines FG, FH, FK, FL, FM are
equal to one another, and the circle described from the
centre F, at the distance of any one of them will pass
through the extremities of the other four ;
and it will touch the straight lines AB, BC, CD, DE, EA,
because the angles at the points G, H, K, L, M are right
angles, [Construction.
and the straight line drawn from the extremity of a dia- "
meter, at right angles to it, touches the circle ; [III. 16.
Therefore each of the straight lines AB, BC, CD, DE, EA
touches the circle.
Wherefore a circle has been inscribed in the given equilateral and equiangular pentagon. q.e.f.
