A history of Greek mathematics - Wilbourhall.org

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NET2EI2 (VERGINGS OR INCLINATIONS) 191
Therefore the triangles BEK, KEG, which have the angle
BEK common, are similar, and
But
Z GBK = Z GKE = Z GEE (from above).
Z HGE = IAGB= Z BCK.
Therefore in the triangles CBK, GHE two angles are
respectively equal, so that
Z GEH — Z GKB also.
But since LGKE = I CHE (from above), K, C, E, E are
concyclic.
Hence
therefore, since
Z CEH+ Z GKE = (two right angles)
Z GEE — Z GKB,
Z GKB + Z Cif# = (two right angles),
and BKE is a straight line.
It is certain, from the nature of this lemma, that Apollonius
made his construction by drawing the circle shown in the
figure.
He would no doubt arrive at it by analysis somewhat as
follows.
Suppose the problem solved, and EK inserted as required
(= h).
Bisect EK in N, and draw NE at right angles to KE
meeting BC produced in E. Draw KM perpendicular to BC,
and produce it to meet AC in L. Then, by the property of
the rhombus, LM = MK, and, since KN = NE also, MN is
parallel to LE.
Now, since the angles at M, N are right, M, K, N, E are
concyclic.
Therefore ICEK = Z.MNK = IGEK, so that C, K, E, E
are concyclic.
Therefore Z BCD = supplement of KCE = LEEK = lEKE,
and the triangles EKE, DGB are similar.
Lastly,
IEBK=IEKE-ICEK=IEEK-ICEK=IEEC=IEKC;
therefore the triangles EBK, EKC are similar, and
BE:EK = EK:EC,
or BE.EC = EK 2 .