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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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