38 CHAPTER 1 Advanced Euclidean Geometry
Proof. Whether or not the
quadrilateral is cyclic, we can con-
struct the point E so that 4 CAD
and 4 CEB are similar. This im-
mediately implies that
CE
CA=
CB
CD
=
BE
DA
,
from which we obtain
BE =CB·DA
CD
. (1.2)
Also, it is clear thatECÂ =BCD̂ ; since also
CD
CA=
CB
CE
,
we may infer that 4 ECA∼4BCD. Therefore,
EA
BD=
CA
CD
,
forcing
EA =
CA·DB
CD
. (1.3)
If it were the case that ABCDwere cyclic, then by (1.1) we would
have
CBÊ +ABĈ = CDÂ +ABĈ = 180◦.But this clearly implies thatA, B,andE are colinear, forcing
EA = AB+BEUsing (1.2) and (1.3) we get
CA·DB
CD