# 164 geometry/bisector.p

## Description

This article is from the Puzzles FAQ,
by Chris Cole chris@questrel.questrel.com and Matthew Daly
mwdaly@pobox.com with numerous contributions by others.

# 164 geometry/bisector.p

Prove if two angle bisectors of a triangle are equal, then the triangle is

isosceles (more specifically, the sides opposite to the two angles

being bisected are equal).

geometry/bisector.s

PROVE: <ABC = <BCA (i.e. triangle ABC is an isosceles triangle)

A
/ \
/ \
D E XP normal to AB
/ \ / \ XQ normal to AC
P /----X----\ Q
/ / \ \
/ / \ \
/ / \ \
B/_______________\C

PROOF :

Let XP and XQ be normals to AC and AB.

Since the three angle bisectors are concurrent, AX bisects angle A

also and therefore XP = XQ.

Let's assume XD > XE.
Then ang(PDX) < ang(QEX)
Now considering triangles BXD and CXE,
the last condition requires that
ang(DBX) > ang(ECX)
OR ang(XBC) > ang(XCB)
OR XC > XB
Thus our assumption leads to :
XC + XD > XE + XB
OR CD > BE
which is a contradiction.

Similarly, one can show that XD < XE leads to a contradiction too.

Hence XD = XE => CX = BX

From which it is easy to prove that the triangle is isosceles.

-- Manish S Prabhu (mprabhu@magnus.acs.ohio-state.edu)

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