3).  With  M  as  centre,  and
radius  MG,  set  off  the  points  H,
K,  and  L,  and  join  GH,  HK,
KL,andLG.  Then  HGLKwill
be  the  square  required.
Fig.  203.
To  construct  a  triangle  similar  to  a  given  triangle  CDE,  and
having  its  perimeter  equal  to  a  given  straight  line  AB.
On  the  given  line  AB  construct
a  triangle  ABF  similar  to  the
given  triangle  CDE,  by  making
the  angles  at  A  and  B  equal  to
the  angles  at  C  and  D  respec-
tively. Bisect  the  angles  at  A
and  B  by  lines  meeting  at  G.
From  G  draw  a  line  parallel  to
FB  till  it  meets  AB  at  L  ;  and
also  a  line  parallel  to  AF  till  it
meets  AB  at  H.  Then  HLG
will  be  the  triangle  required.
CONSTRUCTIOX   OF    SIMILAR   FIGURES.
103


Principles  of  Similitude.
Draw  a  rectangle  ABDC,  and  join  each  angle  to  any
E.  Bisect  EB  in  G,  EA  in  F,  ED  in  K,  and  EC  in  H.
FG,  GK,  KH,  and
HF,  then  FGKH


point
Join


will
be  a  rectangle  with
sides  one -half  the
length  of  the  rectangle
ABCD.  If  we  draw
the  diagonals  BC  and
GH,  we  shall  find
they    are    parallel    to

each    other,  and    that  i"ig-205.
BC  :  GH  as  2:1.  If  we  take  any  point  L- in  BD  and  join
it  to  point  E,  LE  will  intersect  GK  in  M,  and  will  divide  GK
in  the  same  proportion  as  BD  is  divided.
