   @Proposition             Proposition 47
@
   In right-angled triangles the square 
   on the side subtending the right
   angle is equal to the squares on the 
   sides containing the right angle.
   @Proposition
   
   @Figure @
   Let @1 ABC@ be a right-angled triangle@ 
   having the angle BAC right;
   
   I say that 
     @2 the square on BC@
     is equal to 
     @3 the squares on BA, AC@.
   @Figure
   
   @Squares @
   For let there be described on BC the
   @11 square BDEC@, and on BA, AC the
   @12 squares BAGF, ACKH@;
   @Squares
   
   @LineAL @
   Through A let @21 AL@ be drawn 
   parallel to either @22 BD or CE@.
   @LineAL
   
   @Shear1 @
   Now the @31 triangle ABF@ is equal 
   to the @32 triangle CBF@, for they
   have the same @33 base BF@, and are 
   in the same @34 parallels BF, GC@.
   @Shear1
   
   @Angles @
   And, since the @51 angle DBC@ is equal
   to the @52 angle FBA@: for each is right:
   
   Let the @53 angle ABC@ be added to each;
   
   Therefore the whole @54 angle FBC@ is
   equal to the whole @55 angle ABD@.
   @Angles
   
   @Sides @
   Now the @41 side BF@ is equal to the
   @42 side BA@; and the @43 side BC@ is equal
   to the @44 side BD@;
   @Sides
   
   @Rotate @
   Therefore @61 triangle FBC@ is equal
   to @62 triangle ABD@.
   @Rotate
   
   @Shear2 @
   And again the @71 triangle ABD@ is 
   equal to the @72 triangle BDL@, for they
   have the same @73 base BD@, and are 
   in the same @74 parallels BD, AL@.
   @Shear2
   
   @End1 @
   Thus we have shown that @81 triangle ABF@
   is equal to the @82 triangle BDL@.  And
   therefore, the @83 square GB@ is equal
   to the @84 rectangle BL@.
   @End1
   
   @Symmetry @
   And by the same argument, the 
   @91 triangle CAK@ is equal to the
   @92 triangle CEL@.
   
   And the @93 square HC@ is equal to the
   @94 rectangle CL@.
   @Symmetry
   
   @Rehash @
@100   And therefore the square 
   on the side subtending the right
   angle is equal to the squares on the 
   sides containing the right angle.                                          @
   
@101   Which is what was to be proved.                                        @
   @Rehash
