197. Triangles outside of a triangle and concurrence.
by Virgil Nicula, Dec 25, 2010, 5:28 PM
PP1. Outside of the triangle 
construct the isosceles triangles 
in the vertices 
respectively. Prove that the perpendicular
lines from
on 
respectively are concurrently and for any point 
exists the relation 
.
Proof. I"ll use the Carnot's lemma. Denote the projections
of on respectively. Observe that 
and 
. The triangles and 
are orthologically ! Denote 
and for a point obtain that
because 
.
Obtain analogously that
and 
. In conclusion,
.
PP2 (H. Gulicher, Germany). Outside of
construct the right triangles 
and 
, where 
and
. Denote 
so that 
. Prove that 
.
Proof. Denote
and 
. Observe that
and ![$\left\{\begin{array}{cccc}
\blacktriangleright & \frac {MA}{MB}=\frac {[PAC]}{[PBC]}=\frac {AP}{BP}\cdot\frac {AC}{BC}\cdot\frac {\sin\widehat{PAC}}{\sin \widehat{PBC}} & \implies & \frac {MA}{MB}=\frac {1}{\sin x}\cdot\frac ba\cdot\frac {\sin (A+x)}{\cos B}\\\\
\blacktriangleright & \frac {NA}{NC}=\frac {[QAB]}{[QCB]}=\frac {AQ}{CQ}\cdot\frac {AB}{CB}\cdot\frac {\sin\widehat{QAB}}{\sin \widehat{QCB}} & \implies & \frac {NA}{NC}=\frac {1}{\sin y }\cdot\frac ca\cdot\frac {\sin (A+y)}{\cos C}\end{array}\right\|$](//latex.artofproblemsolving.com/0/b/9/0b9b02c51ee9ec5cdb21b76c7630747c6a8cb39d.png)
![$\left(\frac cb\cdot\frac {\cos B}{\cos C}\right)\cdot \left[\sin y\cdot \frac ac\cdot\frac {\cos C}{\sin (A+y)}\right]\cdot$](//latex.artofproblemsolving.com/1/e/3/1e3c28ba93dbf0642cee058157cee7685398ff04.png)
![$\left[\frac {1}{\sin x}\cdot\frac ba\cdot\frac {\sin (A+x)}{\cos B}\right]=$](//latex.artofproblemsolving.com/2/d/a/2daa2e14b113a414f27aa2abccd4da3e705e430d.png)
. Therefore, 
.
See and the previous message.
construct the isosceles triangles 
in the vertices 
respectively. Prove that the perpendicularlines from
on 
respectively are concurrently and for any point 
exists the relation 
.Proof. I"ll use the Carnot's lemma. Denote the projections
of on respectively. Observe that 
and 
. The triangles and 
are orthologically ! Denote 
and for a point obtain that
because 
.Obtain analogously that
and 
. In conclusion,
. PP2 (H. Gulicher, Germany). Outside of
construct the right triangles 
and 
, where 
and
. Denote 
so that 
. Prove that 
.Proof. Denote
and 
. Observe that
and ![$\left\{\begin{array}{cccc}
\blacktriangleright & \frac {MA}{MB}=\frac {[PAC]}{[PBC]}=\frac {AP}{BP}\cdot\frac {AC}{BC}\cdot\frac {\sin\widehat{PAC}}{\sin \widehat{PBC}} & \implies & \frac {MA}{MB}=\frac {1}{\sin x}\cdot\frac ba\cdot\frac {\sin (A+x)}{\cos B}\\\\
\blacktriangleright & \frac {NA}{NC}=\frac {[QAB]}{[QCB]}=\frac {AQ}{CQ}\cdot\frac {AB}{CB}\cdot\frac {\sin\widehat{QAB}}{\sin \widehat{QCB}} & \implies & \frac {NA}{NC}=\frac {1}{\sin y }\cdot\frac ca\cdot\frac {\sin (A+y)}{\cos C}\end{array}\right\|$](http://latex.artofproblemsolving.com/0/b/9/0b9b02c51ee9ec5cdb21b76c7630747c6a8cb39d.png)
![$\left(\frac cb\cdot\frac {\cos B}{\cos C}\right)\cdot \left[\sin y\cdot \frac ac\cdot\frac {\cos C}{\sin (A+y)}\right]\cdot$](http://latex.artofproblemsolving.com/1/e/3/1e3c28ba93dbf0642cee058157cee7685398ff04.png)
![$\left[\frac {1}{\sin x}\cdot\frac ba\cdot\frac {\sin (A+x)}{\cos B}\right]=$](http://latex.artofproblemsolving.com/2/d/a/2daa2e14b113a414f27aa2abccd4da3e705e430d.png)
. Therefore, 
.See and the previous message.
This post has been edited 15 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:29 PM
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