10 CHAPTER 1 Advanced Euclidean Geometry
Proof. Assume that the lines in question are concurrent, meeting in
the pointP. We then have, applying the above lemma three times,
that
1 =
area 4 APC
area 4 BPC·
area 4 APB
area 4 APC·
area 4 BPC
area 4 BPA
=AZ
ZB
·
BX
XC
·
CY
Y A
.
.
To prove the converse we need to
prove that the lines (AX),(BY),
and (CZ) are concurrent, given
that
AZ
ZB·
BX
XC
·
CY
Y Z
= 1.
Let Q = (AX) ∩ (BY), Z′ =
(CQ)∩(AB). Then (AX),(BY),
and (CZ′) are concurrent and so
AZ′
Z′B·
BX
XC
·
CY
Y Z
= 1,
which forces
AZ′
Z′B
=
AZ
ZB
.
This clearly implies thatZ=Z′, proving that the original lines (AX),(BY),
and (CZ) are concurrent.
Menelaus’ theorem is a dual version of Ceva’s theorem and concerns
notlines(i.e., Cevians) but ratherpointson the (extended) edges of