164 geometry/bisector.p
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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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