89. Similarity (parallel/antiparallel).
by Virgil Nicula, Aug 25, 2010, 12:58 AM
Let 
be an interior point for 
. Draw through the lines 
, 
, 
which are 
-parallel / antiparallel-
to sidelines 
, 
, 
respectively
so that
, where 
. Find 
in terms of 
, 
, 
. Answer : 
. See (1) and (2).
Proof (metric). Denote the distancies
, 
, 
of to , , respectively. Observe that ![$\sum\frac {d_a}{h_a}=\sum\frac {ad_a}{ah_a}=\sum \frac {[BPC]}{[ABC]}$](//latex.artofproblemsolving.com/d/8/1/d8181a163597f56dac04a318366dc5a470d2ad43.png)
.
a.s.o. 
a.s.o. 
, i.e. 
. For 
obtain
a.s.o., i.e. has barycentric coordinates 
w.r.t. .
.
Observe that
the quadrilaterals 
, 
, 
are isosceles trapezoids.
Denote midpoints
, 
, 
of ![$[IF]$](//latex.artofproblemsolving.com/7/8/2/782899635cf77445262ce0b52aa0ace7d5e33cc1.png)
, ![$[EH]$](//latex.artofproblemsolving.com/4/4/6/44608dae2c11f0c8867d4419ceb054bd422c19d9.png)
, ![$[GD]$](//latex.artofproblemsolving.com/3/7/6/3763b7d0c97b3389191b653a23a31b58ce56c300.png)
respectively 
.
But
a.s.o. Thus, ![$\left\|\begin{array}{c}
\frac {d_a}{h_a}=\frac {a\left[x(b^2+c^2)-abc\right]}{bc(b^2+c^2-a^2)}\\\\
\frac {d_b}{h_b}=\frac {b\left[x(c^2+a^2)-abc\right]}{ca(c^2+a^2-b^2)}\\\\
\frac {d_c}{h_c}=\frac {c\left[x(a^2+b^2)-abc\right]}{ab(a^2+b^2-c^2)}\end{array}\right\|\ \bigoplus\stackrel{(*)}{\implies}$](//latex.artofproblemsolving.com/9/e/8/9e8839ae8c5b6dd4439793fbae266f03a64ac619.png)
because for 
have
![$\sum\left[u+\frac {u(u+v+w)}{v+w-u}\right]=\sum\frac {2u(v+w)}{v+w-u}$](//latex.artofproblemsolving.com/5/c/9/5c9159aefed21320b3737fd173f99c98599a8120.png)
, what is truly.
For
obtain 
a.s.o., i.e. 
is Lemoine's point of .
Proof (synthetic - luisgeometria's). Assume that , are equal and antiparallel to 
, respectively and 
. Observe that 
is an isosceles trapezoid with 
and from 
obtain that 
is identical to the 
-symmedian of . Thus, point of ''equal antiparallels'' is symmedian point 
of , i.e. 
is concyclic on a circle with center . Let 
be the foot of the -symmedian and be the midpoint of 
, 
are homologous medians of the similar triangles 
. Consequently, we have the following well-known expressions 
.
be an interior point for 
. Draw through the lines 
, 
, 
which are 
-parallel / antiparallel-
to sidelines 
, 
, 
respectivelyso that
, where 
. Find 
in terms of 
, 
, 
. Answer : 
. See (1) and (2).Proof (metric). Denote the distancies
, 
, 
of to , , respectively. Observe that ![$\sum\frac {d_a}{h_a}=\sum\frac {ad_a}{ah_a}=\sum \frac {[BPC]}{[ABC]}$](http://latex.artofproblemsolving.com/d/8/1/d8181a163597f56dac04a318366dc5a470d2ad43.png)
.
a.s.o. 
a.s.o. 
, i.e. 
. For 
obtain
a.s.o., i.e. has barycentric coordinates 
w.r.t. .
.
Observe that
the quadrilaterals 
, 
, 
are isosceles trapezoids.Denote midpoints
, 
, 
of ![$[IF]$](http://latex.artofproblemsolving.com/7/8/2/782899635cf77445262ce0b52aa0ace7d5e33cc1.png)
, ![$[EH]$](http://latex.artofproblemsolving.com/4/4/6/44608dae2c11f0c8867d4419ceb054bd422c19d9.png)
, ![$[GD]$](http://latex.artofproblemsolving.com/3/7/6/3763b7d0c97b3389191b653a23a31b58ce56c300.png)
respectively 
.But
a.s.o. Thus, ![$\left\|\begin{array}{c}
\frac {d_a}{h_a}=\frac {a\left[x(b^2+c^2)-abc\right]}{bc(b^2+c^2-a^2)}\\\\
\frac {d_b}{h_b}=\frac {b\left[x(c^2+a^2)-abc\right]}{ca(c^2+a^2-b^2)}\\\\
\frac {d_c}{h_c}=\frac {c\left[x(a^2+b^2)-abc\right]}{ab(a^2+b^2-c^2)}\end{array}\right\|\ \bigoplus\stackrel{(*)}{\implies}$](http://latex.artofproblemsolving.com/9/e/8/9e8839ae8c5b6dd4439793fbae266f03a64ac619.png)
because for 
have
![$\sum\left[u+\frac {u(u+v+w)}{v+w-u}\right]=\sum\frac {2u(v+w)}{v+w-u}$](http://latex.artofproblemsolving.com/5/c/9/5c9159aefed21320b3737fd173f99c98599a8120.png)
, what is truly.For
obtain 
a.s.o., i.e. 
is Lemoine's point of .Proof (synthetic - luisgeometria's). Assume that
, are equal and antiparallel to 
, respectively and 
. Observe that 
is an isosceles trapezoid with 
and from 
obtain that 
is identical to the 
-symmedian of . Thus, point of ''equal antiparallels'' is symmedian point 
of , i.e. 
is concyclic on a circle with center . Let 
be the foot of the -symmedian and be the midpoint of 
, 
are homologous medians of the similar triangles 
. Consequently, we have the following well-known expressions 
.This post has been edited 55 times. Last edited by Virgil Nicula, Nov 23, 2015, 2:13 PM
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