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Explanation

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Given ΔABC and points X, Y, Z on BC, CA, and AB. O is an arbitrary point. D is the intersection of OA and YZ, E = OB ∩ XZ, F = OC ∩ XY. Then, the cevians AX, BY, CZ in ΔABC concur iff the cevians XD, YE, and ZF in ΔXYZ concur.

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In fact more is true, viz.,

(*)
 AZZB
·
 BXXC
·
 CYYA
=
 ZDDY
·
 YFFX
·
 XEEZ

Note that this is true regardless of the position of point O that does not enter the identity at all.

Thus when one of the expression equals 1 (Ceva's condition), so is the other. This relation has been established earlier at the Cevian Nest page. Here I shall offer another derivation.

The problem as it was formulated above, has been posted in the January 1961 Mathematics Magazine with several ad hoc solutions in the September 1961 issue. What follows is a comment by D. Moody Bailey published in the March-April 1962 issue.

Drop the perpendiculars YG and ZH from Y and Z onto AO. Right triangles DZH and DYG are similar, implying ZD/DY = ZH/YG. On the other hand, from the pair of right triangles AZH and AYG, we obtain ZH = AZ sin(∠OAZ) and YG = YA sin(∠OAY). Substitution then yields

 ZDDY
=
 AZYA
·
 sin∠OAZsin∠OAY

and, in a similar fashion,

 YFFX
=
 CYXC
·
 sin∠OCYsin∠OCX
and
 XEEZ
=
 BXZB
·
 sin∠OBXsin∠OBZ

Consequently,

 AZZB
·
 BXXC
·
 CYYA
=
 ZDDY
·
 YFFX
·
 XEEZ
·(
 sin∠OAYsin∠OAZ
·
 sin∠OBZsin∠OBX
·
 sin∠OCXsin∠OCY
)

However, the expression in the parentheses equals 1 due to the Trigonometric form of Ceva's theorem applied in ΔABC and the Cevians through point O. Thus (*) holds precisely because the three cevians AO, BO, CO meet in a point (O) so that, perhaps less surprisingly now, it holds regardless of the specific location of O.

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