And because FB is equal to FC,
and FK is common to the two triangles BFK, CFK ;
the two sides BF, FK are equal to the two sides CF, FK,
each to each ;
and the base BK was shewn equal to the base CK;
therefore the angle BFK is equal to the angle CFK, [I. 8.
and the angle BKF to the angle CKF. [I. 4.
Therefore the angle BFC is double of the angle CFK,
and the angle BKC is double of the angle CKF.
For the same reason the angle CFD is double of the angle CFL, and the angle CLD is double of the angle CLF.
And because the arc BC is equal to the arc CD,
the angle BFC is equal to the angle CFD ; [III. 27.
and the angle BFC is double of the angle CFK, and the
angle CFD is double of the angle CFL ;
therefore the angle CFK is equal to the angle CFL. [Ax. 7.
And the right angle FCK is equal to the right angle FCL.
Therefore in the two triangles FCK, FCL, there are two
angles of the one equal to two angles of the other, each to
each ;
and the side FC, which is adjacent to the equal angles in
each, is common to both ;
therefore their other sides are equal, each to each, and the
third angle of the one equal to the third angle of the other ;
therefore the straight line CK is equal to the straight line
CL, and the angle FKC to the angle FLC. [I. 26.
And because CK is equal to CL, LK is double of CK.
In the same manner it may be shewn that HK is double of BK.
And because BK is equal to CK, as was shewn,
and that HK is double of BK, and LK double of CK,
therefore HK is equal to LK. [Axiom 6.
In the same manner it may be shewn that GH, txM, ML are each of them equal to HK or LK; therefore the pentagon GHKLM is equilateral.
It is also equiangular.
