ABCD is a rhombus. Take points E, F, G, H on sides AB, BC, CD, DA respectively so that EF and GH are tangent to the incircle of ABCD. Show that EH and FG are parallel.
It is easy to show that EH and FG are parallel if
and only if triangles HAE and FCG are similar.
Let O be the intersection of diagonals AC and BD
(also, it's the center of the incircle). Let E',
F', G', and H' be the points of tangency of the
incircle with the sides AB, BC, CD, and DA
respectively. Let I and J be the points of
tangency of the incircle with EF and GH
respectively. In the following a string of three
capital letter denotes an angle unless specified
otherwise. Let the twelve angles about O be
u = H'OH = HOJ
v = JOG = GOG'
z = G'OC = COF'
x = F'OF = FOI
y = IOE = EOE'
z = E'OA = AOH'.
Clearly, the following holds
u + v + z = x + y + z = 90. (1)
With algebra and a trig. identity (1) implies
tan(x)+tan(z) tan(v)+tan(z)
--------------- = ---------------. (2)
tan(z)+tan(u) tan(z)+tan(y)
Multiplying both sides of (2) with r/r (where
r is radius of the incircle) gives
r*tan(x)+r*tan(z) r*tan(v)+r*tan(z)
------------------- = -------------------
r*tan(z)+r*tan(u) r*tan(z)+r*tan(y)
or
|EE'| + |E'A| |GG'| + |G'C|
--------------- = ---------------
|AH'| + |H'H| |CF'| + |F'F|
or
|EA| |GC|
------ = ------.
|AH| |CF|
This and HAE = FCG implies the triangles
are similar.
QED
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Posted by Bractals
on 2013-01-14 13:07:32 |
Solution