﻿Solution 
  to 
  an 
  Historical 
  Theorem 
  in 
  Geometry. 
  985 
  

  

  The 
  original 
  theorem 
  was, 
  " 
  If 
  the 
  bisectors 
  of 
  the 
  base 
  

   angles 
  of 
  a 
  triangle 
  are 
  equal, 
  the 
  triangle 
  is 
  isosceles." 
  It 
  

   follows 
  as 
  a 
  corollary 
  to 
  the 
  following 
  : 
  — 
  

  

  Through 
  any 
  point 
  in 
  the 
  bisector 
  of 
  an 
  angle, 
  two, 
  and 
  

   only 
  two 
  equal 
  straight 
  lines, 
  terminated 
  by 
  the 
  arms 
  of 
  the 
  

   angle, 
  may 
  be 
  drawn 
  and 
  the 
  segments 
  into 
  which 
  the 
  lines 
  

   are 
  divided 
  at 
  the 
  point 
  are 
  equal, 
  each 
  to 
  each. 
  

  

  Let 
  ACB 
  be 
  any 
  angle 
  and 
  let 
  CO 
  be 
  its 
  bisector. 
  

  

  It 
  is 
  required 
  to 
  prove 
  that 
  through 
  any 
  point 
  P 
  in 
  CO, 
  

   two 
  and 
  only 
  two 
  equal 
  straight 
  lines 
  terminated 
  by 
  the 
  arms 
  

   CA, 
  CB 
  of 
  the 
  angle 
  may 
  be 
  drawn. 
  

  

  Through 
  P 
  draw 
  any 
  pair 
  of 
  lines 
  XPY, 
  WPZ 
  in 
  such 
  a 
  

   manner 
  that 
  

  

  WPO=XPC. 
  

   It 
  is 
  obvious 
  by 
  I. 
  26 
  that 
  

  

  AXPCeeAWPC 
  .-. 
  XP 
  = 
  WP, 
  

   AYPCeeAZPG 
  .-. 
  PY=PZ. 
  

   XP 
  + 
  PY=WP 
  + 
  PZ, 
  i.e., 
  XY 
  = 
  WZ. 
  

   CXP=CWP 
  (v 
  AXPCeeAWPC), 
  

   Z, 
  X, 
  W, 
  Y 
  are 
  eoncyclic. 
  

   Also 
  A 
  ZPXeeAYPW 
  (1.4) 
  .-. 
  ZX=YW. 
  

  

  These 
  are 
  equal 
  chords 
  of 
  the 
  0ZXWY 
  and 
  are 
  therefore 
  

   equidistant 
  from 
  its 
  centre: 
  

  

  ,\ 
  the 
  centre 
  must 
  Jie 
  in 
  the 
  bisector 
  CO, 
  for 
  CO 
  is 
  the 
  locus 
  

   of 
  points 
  equidistant 
  from 
  CA 
  and 
  CB. 
  

  

  Let 
  the 
  perpendicular 
  bisector 
  of 
  XZ 
  (or 
  WY) 
  cut 
  CO 
  in 
  

   Q, 
  Q 
  is 
  the 
  centre 
  of 
  ©ZXWY. 
  

  

  Draw 
  the 
  circle 
  and 
  through 
  P 
  draw 
  any 
  chord 
  MN 
  cutting 
  

   CA, 
  CB 
  in 
  R 
  and 
  S 
  respectively. 
  

  

  Phil. 
  Mag. 
  S. 
  6. 
  Vol. 
  26. 
  No. 
  156. 
  Dec. 
  1913. 
  3 
  X 
  - 
  

  

  likewise 
  

   By 
  addition 
  

   Again, 
  since 
  

  

  