202. Concurrent lines in a right triangle.
by Virgil Nicula, Dec 31, 2010, 8:27 AM
Proposed problem. Let
be an
-right triangle. For a point
denote the reflection
of 
in the line
and the point
for which
. Prove that
.
Proof 1.
Proof 2.
Indeed,
the division
is harmonically.
Proof 3.
Proof.





in the line




Proof 1.
Proof 1 (with harmonical division). Denote
. Thus,

. From the
-right triangle
obtain that
. In conclusion,
.
Since
and
obtain that the division
is harmonically
.








Since





Proof 2.
Proof 2 (with perspective). Denote as
,
,
,
,
. From the complete quadrilateral
conclude that the points
,
are harmonic conjugates w.r.t. the points
,
. Since
,
and
obtain that 
is cyclic ( taken as diameter
) and the line
is the bisector of
because
. Since
obtain that
, the external angle
bisector of
through vertex
. Because the points
,
,
are collinear, based on the Desarques’ theorem obtain that
and 
are perspective
,
and
. Hence
.
Remark (a short commentary). The proposed problem is a very nice application of the following remarkable harmonical division














is cyclic ( taken as diameter
![$[BD]$](http://latex.artofproblemsolving.com/3/2/6/3261e689901bce018ecdef47b9bc60a78ead3746.png)





bisector of







are perspective




Remark (a short commentary). The proposed problem is a very nice application of the following remarkable harmonical division
Quote:
Lemma. Let
,
be two secant circles in
so that
. Consider a point
and a line
for which
.
Denote
and
so that
separates
,
. Then the division
is harmonically.
Particular case. Let
be a triangle with the orthocenter
. Denote
and the intersections
, 
between the line
and the circle with the diameter
. Then the division
is harmonically.







Denote






Particular case. Let





between the line

![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)

Indeed,


Proof 3.
Proof 3 (with harmonical division). Denote
and
. Apply upper particular case to
with the orthocenter
and obtain that the division
is harmonically
.
The following interesting problem is a nice consequence of the harmonical division which was mentioned in the above lemma :







The following interesting problem is a nice consequence of the harmonical division which was mentioned in the above lemma :
Quote:
Proposed problem. Let
,
be two secant circles in
so that
. Consider a point
and a line
for which
. Denote
and
so that
separates
,
. For
(line !) denote
.Then
.















Proof.
Proof. From the upper lemma obtain that the division
is harmonically, i.e.
. Apply the Menelaus' theorem
to the transversal
and the triangle
, i.e.
.


to the transversal





This post has been edited 19 times. Last edited by Virgil Nicula, Nov 22, 2015, 5:12 PM