SECTION 1.2 Triangle Geometry 15
property of the incenter will be given in
Exercise 12 on page 153.) However, we shall proceed below to give
another proof of this fact, based on Ceva’s Theorem.
Proof that the angle bisectors of 4 ABC are concurrent. In
order to accomplish this, we shall first prove the
Angle Bisector Theorem. We
are given the triangle 4 ABC with
line segment[BP] (as indicated to
the right). Then
AB
BC=
AP
PC
⇔ABP̂ =PBC.̂
Proof (⇐). We drop altitudes
fromP to (AB) and (BC); call the
points so determinedZ andY, re-
spectively. Drop an altitude from
B to (AC) and call the resulting
point X. Clearly PZ = PY as
4 PZB∼= 4 PY B. Next, we have
4 ABX∼4APZ⇒
AB
AP
=
BX
PZ
=
BX
PY
.
Likewise,
4 CBX ∼4CPY ⇒
CB
CP
=
BX
PY
.
Therefore,
AB
BC