316. Some properties of isogonal rays in an angle of ABC.
by Virgil Nicula, Sep 8, 2011, 11:31 PM
PP1. Let 
be a triangle with circumcircle 
. Denote the midpoint 
of ![$[BC]$](//latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and define the tangent 
in 
to the circle . Thus, denote 
, 
and the symmedian 
where 
. Prove that :
, i.e. 
is an harmonic division; 
and 
is tangent in 
to 
, i.e. 
.
is harmonic quadrilateral, where 
, i.e. 
and 
is an harmonic division.
Proof.
. From the Steiner's relation
obtain that
, i.e. is an harmonic division. Suppose w.l.o.g. 
. Since 
obtain
, i.e. 
. Denote 
. Observe that 
,
i.e.
, i.e. is tangent in to and . Since 
obtain that 
and 
, where 
and 
. Denote
. I"ll show that 
. Indeed, 
, i.e. 
is an harmonic quadrilateral.
Observe that
, i.e. 
and 
, i.e. the division is harmonically.
PP2. Let be a triangle with the incenter 
and the circumcircle 
. Consider two points
so that the ray 
is the bisector of the angle 
. The lines 
, 
meet again
in the points
, 
respectively. Prove that 
.
Study some particular cases, for example the remarkable pairs of the isogonal points
, 
and 
, where
- incenter, 
- orthocenter, 
- circumcenter, 
- centroid and 
- Lemoin's point (symmedian center) for the triangle .
Proof.
, i.e. 
and 
, i.e. the relation 
. Using the power of 
w.r.t. , i.e. 
and
the Stewart's relation
obtain that 
. In conclusion, we have
, i.e. the relation 
and for 
obtain that 
.
Some remarkable particular cases.
. In this case 
, 
and 
. Thus, 
and 
. From 
obtain that 
, i.e. 
.
. In this case, 
, 
, 
, 
. Thus, the first relation becomes
. From here obtain ![$S=[ABC]=\frac {ah_a}{2}\iff$](//latex.artofproblemsolving.com/b/7/4/b74ee9840b969d2396d68f3fa739772f66b6cba9.png)
.
. In this case 
, 
, 
, 
. Thus, the second relation becomes 
.
PP3. Let be a triangle with the incenter . Denote 
, 
,
and 
, 
. Prove that 
.
Proof. Using an well-known relation obtain that
. Denote 
and 
.
Using Menelaus' theorem for the transversals in the mentioned triangles
. From the Steiner's theorem obtain that 
, i.e. .
PP4. Let
be a triangle with the circumcircle 
. Denote the following points : the middlepoint 
of the side ![$ [BC]$](//latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
; the point 
for which 
; the point 
for which
; the 
-symmedian 
, where 
; the point 
for which 
. Then 
, i.e. the quadrilateral 
is cyclically.
Proof. Let
- the tangent in 
to and 
. Thus, 
and 
. From the first well-known property, 
and 
is the-symmedian in 
. From the second well-known property, the division 
is harmonically. Since 
results
. Show easily that 
(a.a.) . Thus, 
, i.e. 
. From the third well-known property 
get
, i.e. 
. Since 
obtain 
and 
. Since and 
obtain
, i.e. the quadrilateral is cyclically. In conclusion, .
Remark. Here are another interesting metrical relations :
.
be a triangle with circumcircle 
. Denote the midpoint 
of ![$[BC]$](http://latex.artofproblemsolving.com/e/a/1/ea1d44f3905940ec53e7eebd2aa5e491eb9e3732.png)
and define the tangent 
in 
to the circle
. Thus, denote 
, 
and the symmedian 
where 
. Prove that :
, i.e. 
is an harmonic division; 
and 
is tangent in 
to 
, i.e. 
.
is harmonic quadrilateral, where 
, i.e. 
and 
is an harmonic division.Proof.
. From the Steiner's relationobtain that
, i.e. is an harmonic division. Suppose w.l.o.g. 
. Since 
obtain
, i.e. 
. Denote 
. Observe that 
,i.e.
, i.e. is tangent in to and . Since 
obtain that 
and 
, where 
and 
. Denote
. I"ll show that 
. Indeed, 
, i.e. 
is an harmonic quadrilateral.Observe that
, i.e. 
and 
, i.e. the division is harmonically.PP2. Let
be a triangle with the incenter 
and the circumcircle 
. Consider two points
so that the ray 
is the bisector of the angle 
. The lines 
, 
meet again in the points
, 
respectively. Prove that 
.Study some particular cases, for example the remarkable pairs of the isogonal points
, 
and 
, where - incenter, 
- orthocenter, 
- circumcenter, 
- centroid and 
- Lemoin's point (symmedian center) for the triangle . Proof.
, i.e. 
and 
, i.e. the relation 
. Using the power of 
w.r.t. , i.e. 
andthe Stewart's relation
obtain that 
. In conclusion, we have
, i.e. the relation 
and for 
obtain that 
.Some remarkable particular cases.
. In this case 
, 
and 
. Thus, 
and 
. From 
obtain that 
, i.e. 
.
. In this case, 
, 
, 
, 
. Thus, the first relation becomes
. From here obtain ![$S=[ABC]=\frac {ah_a}{2}\iff$](http://latex.artofproblemsolving.com/b/7/4/b74ee9840b969d2396d68f3fa739772f66b6cba9.png)
.
. In this case 
, 
, 
, 
. Thus, the second relation becomes 
. PP3. Let
be a triangle with the incenter . Denote 
, 
,
and 
, 
. Prove that 
.Proof. Using an well-known relation obtain that
. Denote 
and 
.Using Menelaus' theorem for the transversals in the mentioned triangles
. From the Steiner's theorem obtain that 
, i.e. .PP4. Let
be a triangle with the circumcircle 
. Denote the following points : the middlepoint 
of the side ![$ [BC]$](http://latex.artofproblemsolving.com/3/5/5/3550468aa97af843ef34b8868728963dec043efe.png)
; the point 
for which 
; the point 
for which
; the 
-symmedian 
, where 
; the point 
for which 
. Then 
, i.e. the quadrilateral 
is cyclically.Proof. Let
- the tangent in 
to and 
. Thus, 
and 
. From the first well-known property, 
and 
is the
-symmedian in 
. From the second well-known property, the division 
is harmonically. Since 
results
. Show easily that 
(a.a.) . Thus, 
, i.e. 
. From the third well-known property 
get
, i.e. 
. Since 
obtain 
and 
. Since and 
obtain
, i.e. the quadrilateral is cyclically. In conclusion, .Remark. Here are another interesting metrical relations :
. This post has been edited 62 times. Last edited by Virgil Nicula, Nov 25, 2016, 6:32 PM
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