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

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                       A
                      / 
                     /   
                    D     E            XP normal to AB
                   /    /            XQ normal to AC
                P /----X---- Q
                 /   /      
                /  /         
               / /            
              B/'' '''''' '''' '' '''' '''' '''''' _C
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PROOF :
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  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)
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OR     ang(XBC) > ang(XCB)
OR        XC    >  XB

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   Thus our assumption leads to :
        XC + XD >  XE + XB
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OR         CD  > BE
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  which is a contradiction.


  Similarly, one can show that XD < XE leads to a contradiction too.
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Hence  XD <code> XE  </code>> 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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