 # 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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