Advanced High-School Mathematics

12 CHAPTER 1 Advanced Euclidean Geometry

Case 1. We assume first that
X, Y, andZare collinear and drop
altitudesh 1 , h 2 ,andh 3 as indicated
in the figure to the right. Using ob-
vious similar triangles, we get

AZ

ZB

= +

h 1

h 2

;

BX

XC

= +

h 2

h 3

;

CY

Y A

=−

h 3

h 1

,

in which case we clearly obtain

AZ

ZB

×

BX

XC

×

CY

Y A

=− 1.

To prove the converse, we may assume thatX is on [BC], Zis on
[AB], and thatY is on (AC) with AZZB·BXXC·CYY A=− 1 .We letX′be the
intersection of (ZY) with [BC] and infer from the above that

AZ

ZB

·

BX′

X′C

·

CY

Y A

=− 1.

It follows that BXXC =BX
′
X′C, from which we infer easily thatX=X

′,and

soX, Y, andZare collinear.

Case 2. Again, we drop altitudes from
A,B, andC and use obvious similar tri-
angles, to get

AZ

ZB

=−

h 1

h 2

;

BX

XC

=−

h 2

h 3

;

AY

Y C

=−

h 1

h 3

;

it follows immediately that

AZ

ZB

·

BX

XC

·

CY

Y A

=− 1.

The converse is proved exactly as above.