51. A nice problem with perpendicularity from Jaime.
by Virgil Nicula, Jul 6, 2010, 1:41 PM
Quote:
In an 
-right 
denote the midpoint 
of the side ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the foot 
of the -altitude and the point 
for which 
. For a point 
denote the intersection 
. Prove that 
.
-right 
denote the midpoint 
of the side ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
, the foot 
of the -altitude and the point 
for which 
. For a point 
denote the intersection 
. Prove that 
.Proof 1. Denote
and 
. Observe that 
and 
. Since 
and 
obtain that the points 
, 
are omologously in this similarity. Since 
, 
and 
results that we can obtain the triangle 
from the triangle 
by a rotation with 
and a homothety. In conclusion, .Proof 2 (directly, without geometrical transformations - rotation and homothety). Denote
and . Observe that . Since and obtain that the points , are omologously in this similarity. Therefore, 
. Since and obtain 
, i.e. .Quote:
Generalization. In an -right choose a point on the line of -altitude and a point 
on the line of -median. Denote for which , 
, and 
. Prove that .
obtain the proposed problem by Jaime from hereProof. Denote
. Observe that . Since and obtain that the points , are omologously in this similarity. Therefore, . Since and obtain , i.e. .Quote:
Equivalent enunciation. Let 
be a -right triangle. Choose a point 
on the sideline 
and denote the midpoint of ![$[CF]$](//latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
. Denote 
so that 
and for which . Choose a point 
on the line 
and denote 
and 
. Prove that 
.
be a -right triangle. Choose a point 
on the sideline 
and denote the midpoint of ![$[CF]$](http://latex.artofproblemsolving.com/2/1/b/21bdd766ec1757878aeae83d55b7ab5917af4537.png)
. Denote so that 
and for which . Choose a point 
on the line 
and denote 
and 
. Prove that 
.Quote:
Generalization. In with 
consider the point 
so that 
, the point 
so that 
, the point 
for which 
. For a point 
define 
and 
. Prove that the quadrilateral 
is cyclically . See here.Proof. Denote
. I"ll show the relation 
. Thus, 
.
.Observe that
, i.e. 
. Apply the Menelaus' theorem to the transversal 
and 
. Apply in 
an well-known relation 
. Therefore, 
, 
, what is truly. Therefore, 
and for the points 
, 
exists the relation 
. It means these points are omologously in similarity. Thus, 
, i.e. the quadrilateral is cyclically.Remark. Can define otherwise the point
from the relation 
. Consider the point 
so that 
. Observe that 
and 
, i.e. 
what means that 
, i.e. 
the rays 
, 
are isogonally in the angle 
.A similar proposed problem. In
consider the point so that 
. A circle through 
, cut , in 
, 
respectively. Denote the midpoints 
, 
of ![$[AB]$](http://latex.artofproblemsolving.com/a/d/a/ada6f54288b7a2cdd299eba0055f8c8d19916b4b.png)
, ![$[NP]$](http://latex.artofproblemsolving.com/4/b/f/4bf4a2a9ab593d9ee3bd6f3316f38ad187f4cbb9.png)
respectively and 
. Prove that the quadrilateral 
is cyclically. Remark. Another similar problem is hereThis post has been edited 22 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:44 PM
August 2018
January 2018