Re: Tangential quadrilateral well known properties?
(QFG message 26)
Mosca Sebastiano:
ABCD is a circumscribed quadrilateral,
PQRS the points of tangency with the inscribed circle center O,
M and N are the midpoints of AB and CD,
K is the intersection of SQ and MN
Prove OK is parallel to PR.
C
D
R
N
Q
K
S
O
A
P
B
M
Here is a simple calculation with barycentric coordinates, using
the QA-Diagonal Triangle of the quadrigon PQRS as reference
triangle:
Let A0B0C0 be the reference triangle for barycentric coordinates,
obtuse-angled at B0.
Use the polar circle ([1], p.37) of the reference triangle with the
equation
S A x ²  S B y ²  SC z ²  0
for a cyclic quadrigon PQRS:
P(u : v : w), Q(u : v : w), R(u : v : w), S (u : v : w)
S u ²  SC w²
with v ²  A
.
 SB
Co
Ao
A
M
B
P
Q
Bo
K
S
O
D
R
N
C
The tangents in P, Q, R, S have the intersections
A(S B v : S Au : 0), B(0 : SC w : S B v),
C (S B v : S Au : 0), D(0 : SC w : S B v) .
The midpoints of AB and CD are
M (S B v(S B v  SC w) : S A S Buv  2S A SC uw  S B SC vw : S B v(S B v  S Au))
N (S B v(S B v  SC w) : S A S Buv  2S A SC uw  S B SC vw: S B v(S B v  S Au)) .
This gives the intersection of QS and MN
S S u ²  ( S ²  S A SC )uw  S B SC w²
K (u : A B
: w) .
S B ( S Au  SC w)
The center of the polar circle is the orthocenter of the reference
triangle O(S B SC :S C S A : S A S B ) .
The common point at infinity of KO and PR is (u : u  w : w) ,
so KO and PR are parallel.
[1]
http://forumgeom.fau.edu/FG2004volume4/FG200405.pdf
Eckart Schmidt
http://eckartschmidt.de
[email protected]