. REFER TO THE DIAGRAM. QUADRILATERAL ABCD WITH DIAGONALS BD...
1..
Refer to the diagram. Quadrilateral ABCD with diagonals
BD
and
E
G
AC bounds the rhombus
EFGH
as shown in the figure. If
BD
k
AC
=
,
B
F
C
ABCD
||
EF
AC
, and
FG BD
||
, find
Area
Area
EFGH
.
Solution:
AE
, then
EH
BD
BE
=
1
−
λ
,
=
λ
,
=
λ
. Similarly,
EF
=
(
1
−
λ
)
AC
, but EH=EF, so
EH
Let
=
λ
BD
AB
BA
=
−
k
BD
, then
=
1
λ
.
that
λ
⋅
BD
=
(
1
−
λ
)
⋅
AC
and
λ
λ
kH
AC
1
ϑ
=
⋅
⋅
2
sin
+
2
S
k
k
1
1
=
⋅
=
+
.
Hence
(
) (
)
λ
2
EF
sin
k
EFGH
+
+