99. Some simple and nice geometry problems.
by Virgil Nicula, Sep 6, 2010, 2:58 AM
http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=365571
Let 
, 
be two isosceles triangles so that 
, 
and 
, 
.
Denote
, 
. Prove that 
is one from the bisectors of 
.
Proof. Denote
, 
, 
. Apply Menelaus' theorem to transversals 
, 
in triangles 
, 
respectively :
. Apply Ceva's theorem to the point 
and 
. In conclusion, 
ray is one from the bisectors of .
Consider in 
the 
-median 
, the -bisector 
, where 
and the point 
for which 
. Prove that 
.
Proof 1 (metric). Suppose w.l.o.g.
. Prove easily that 
and 
. Thus 
. Denote 
. Apply Menelaus' theorem to transversal 
in 
, i.e. 
.In conclusion, the triangle 
is -isosceles .
Proof 2 (synthetic). Denote
and the midpoints 
, 
of ![$[KD]$](//latex.artofproblemsolving.com/5/d/3/5d3f8adbb8bcc9de2fc03de544ec69ad309ad137.png)
, ![$[AC]$](//latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively. Therefore, the points 
, , 
,
are colinearly ,
and the triangle 
is -isosceles because 
. Since is midpoint of obtain 
.
Let 
be a 
- right triangle . Denote ![$ \left|\begin{array}{c} H\in BC\ ,\ AH\perp BC \\
\\
N\in [AB]\ ,\ NA = NB \\
\\
M\in [AC]\ ,\ MA = MC\end{array}\right|$](//latex.artofproblemsolving.com/6/5/3/653ea2d4ddb0e5b122e56ef3cc7c6f2327384f1b.png)
and 
. Prove that 
.
Proof 1 (synthetic). Let
and 
. Since 
, 
belong to the medians 
, 
respectively obtain :
.
Proof 2 (metric). Denote
, 
. Apply the Ceva's theorem to the points
, 
in 
.
Proof 3 (metric). Apply the Menelaus' theorem to the transversal
and the triangle 
. Apply an well-known relation to the cevian 
in the triangle 
. Thus, 
. Analogously obtain 
. In conclusion, . Shortly, 
.
Remark. Construct, outside of
, the rectangle 
, where 
. Denote the circumcircle 
of this rectangle. Prove easily that 
, 
, 
, i.e. the points 
, 
belong to the circle . If 
, then 
and 
.
Let 
, 
be two isosceles triangles so that 
, 
and 
, 
.Denote
, 
. Prove that 
is one from the bisectors of 
.Proof. Denote
, 
, 
. Apply Menelaus' theorem to transversals 
, 
in triangles 
, 
respectively :
. Apply Ceva's theorem to the point 
and 
. In conclusion, 
ray is one from the bisectors of .
Consider in 
the 
-median 
, the -bisector 
, where 
and the point 
for which 
. Prove that 
.Proof 1 (metric). Suppose w.l.o.g.
. Prove easily that 
and 
. Thus 
. Denote 
. Apply Menelaus' theorem to transversal 
in 
, i.e. 
.In conclusion, the triangle 
is -isosceles .Proof 2 (synthetic). Denote
and the midpoints 
, 
of ![$[KD]$](http://latex.artofproblemsolving.com/5/d/3/5d3f8adbb8bcc9de2fc03de544ec69ad309ad137.png)
, ![$[AC]$](http://latex.artofproblemsolving.com/0/9/3/0936990e6625d65357ca51006c08c9fe3e04ba0c.png)
respectively. Therefore, the points 
, , 
, are colinearly ,
and the triangle 
is -isosceles because 
. Since is midpoint of obtain 
. 
Let 
be a 
- right triangle . Denote ![$ \left|\begin{array}{c} H\in BC\ ,\ AH\perp BC \\
\\
N\in [AB]\ ,\ NA = NB \\
\\
M\in [AC]\ ,\ MA = MC\end{array}\right|$](http://latex.artofproblemsolving.com/6/5/3/653ea2d4ddb0e5b122e56ef3cc7c6f2327384f1b.png)
and 
. Prove that 
.Proof 1 (synthetic). Let
and 
. Since 
, 
belong to the medians 
, 
respectively obtain :
.Proof 2 (metric). Denote
, 
. Apply the Ceva's theorem to the points
, 
in 
.Proof 3 (metric). Apply the Menelaus' theorem to the transversal
and the triangle 
. Apply an well-known relation to the cevian 
in the triangle 
. Thus, 
. Analogously obtain 
. In conclusion, . Shortly, 
.Remark. Construct, outside of
, the rectangle 
, where 
. Denote the circumcircle 
of this rectangle. Prove easily that 
, 
, 
, i.e. the points 
, 
belong to the circle . If 
, then 
and 
.This post has been edited 88 times. Last edited by Virgil Nicula, Nov 23, 2015, 8:32 AM
August 2018
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