16 CHAPTER 1 Advanced Euclidean Geometry
(⇒). Here we’re given thatABBC =APPC.Let
P′be the point determined by the angle
bisector (BP′) of ABĈ. Then by what
has already been proved above, we have
AP
BC =
AP′
P′C.But this implies that
AP
PC=
AP′
P′C
⇒P =P′.
Conclusion of the proof that angle bisectors are concurrent.
First of all, it is clear that the relevant ratios are all positive. By the
Angle Bisector Theorem,
AB
BC=
AY
Y C
,
BC
CA
=
BZ
ZA
,
AB
AC
=
BX
XC
;
therefore,
AZ
BZ×
BX
XC
×
CY
Y A
=
CA
BC
×
AB
AC
×
BC
AB
= 1.
Ceva’s theorem now finishes the job!
Exercises
- The Angle Bisector Theorem involved the bisection of one of the
given triangle’sinteriorangles. Now letP be a point on the line
(AC) external to the segment [AC]. Show that the line (BP)
bisects the external angle atBif and only if
AB
BC=
AP
PC
.
- You are given the triangle 4 ABC. Let X be the point of inter-
section of the bisector ofBAĈ with [BC] and letY be the point
of intersection of the bisector ofCBÂ with [AC]. Finally, letZbe
the point of intersection of theexterior angle bisector at C with
the line (AB). Show thatX, Y, andZare colinear.^3
(^3) What happens if the exterior angle bisector atCis parallel with (AB)?