Table of contents
1. Introduction
2. Isogonal lines – what is it
3. Isogonal lines theory
4. Proof of the isogonal lines theory
5. Inverse theory
6. “Why is it theory important in study?”
7. Conclusion
8. Bibliography
9. Supplements
Introduction
“Just enplane me…
But not simply explain but it must be clean at the same time”
Grishcovech
I am going to have my first state exams next year to choose what
I
will study; in what institute and what profession I want to get.
The total list of professions is very large because of development
of
technology and science. A lot of new spheres appear. The main universities control
and react immediately their appearing. Offering studying in these specialties. Math
is the base of most of them. We have some time to think, discuss, test yourself.
I was given a very unusual reference book for geometry by Arseniy Akopyan. You
can find theories of classical geometry in pictures. Looking through and testing
yourself on example 1.9 I had to realize that I don’t know this theory.
Isogonal lines.
Two straight lines passing through the vertex of an angle and making equal angles with
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its bisector are called isogonal straight lines (isogonals) relative to this angle or relative
to the sides of this angle. Isogonal lines are also sometimes called inverse lines.
Let ℓ bisectrix ∠АВС, and the rays in BF and BE form equal angles
with the bisectrix. BF and BE – isogonals ∠АВС. Obviously, ∠1 = ∠2.
Ptolemy, Brahmaputra, Lanrange, Steiner and many others studied
isogonal lines.
Theory In triangle ABC, the height and radius of the circumscribed circle, drawn
from the same vertex, are isogonal.
Proof. BD is the diameter of the circle, so the triangle
DCB is right-angle, CDB and CAB are based on the arc CD
and are therefore equal. And therefore, having considered two
right-angled triangles DCB and BHA with corners CDB = CAB, we conclude that HBA
= CBD and the diameter of the isogonals is equal the the height of the triangle.
This theory has an inverse theory. In order for line to be isogonal to
the height of a triangle, it is necessary that the center of the
circumscribed belong to it.
Why is it theory important in study with it you can easily solve
problems below.
Problem 1. The height and median of a triangle drawn to one side divide the angle
into three equal parts. Find the angle of the triangle.
Solution. BH and BM are isogonal, so the center of the circumscribed circle
belongs to the median, but the bisector also belongs to the
median, so the center of the circumscribed circle and the point
M coincide. CM-MA=BM are the radii of the circumscribed
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circle, and the angle is then angle CBA = 90 CBM=MBH=HBA=30 according to
the problem, therefore BAC=90-30=60, BCA=30
Problem 2. Prove that the line Ov, where O is the center of the circumscribed
circle of triangle ABC, is perpendicular to the segment nan,
where H2Hy is the base of the height drawn from vertices A and
C.
Solution. Consider an acute triangle, ABC, BO
– the radius of the circle. BD – diameter, BD and BH1,
are isogonal, so CBD=ABH1 quadrilateral CH2H3A
can be circumscribed by a circle with diameter CA, so
CH2A=CH3A=90, CH2H3+H3H2B=180, hence H3H2B = H3AC =
BAC. In triangle ABH and H2BT BHA = 90 ∠Н3Н2В =∠ВАС,
∠ Н2ВD=∠АВН1
∠ ВН1А=90 , consequently ∠Н2 В D=90
Problem 3. The height, median and bisector of s triangle drawn to one side divide
the angle into four equal parts. Find the angles of the triangle.
Solution. The median and height of the isogonal, therefore, the
median lies on the diagonal, and since the bisector also lies on
the circumscribed circle will coincide with the point M.
Therefore, the angles ABM=BAM=MBL=LBH=HBC=a. From right-angle triangle
ABH find BAM=90: 4a=22,5 FDC=4a=90. DCF=90-22,5=67,5
Problem 4. Arbitrary points X, XQ and XT are taken inside the angle ABC –
perpendiculars and sides of the angle. BK perpendicular to QT. Proof that
QBK=TBX
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Solution. Near quadrilateral TBOX one can circumscribed a circle with
diameter BX (the angles are opposite to right angles), so BX and VC
and triangle VTO, and hence with QBK=NBX.
Problem 5. Quadrilateral ABCD is circumscribed in a circle with center O, angle
BAO is equal to angle DAC. Prove that the diagonals of a quadrilateral are
perpendicular.
Solution. AC and AO are isogonal for triangle BAD and, therefore,
if AO is a line containing the center of the circumscribed circle,
then AC is the altitude. AC OBD.
Conclusion
Studying the literature and internet resources, I realized that equiangular right angles are
an important theory for planimetry and a convenient way to solve problems of this type.
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There are many more theories in this theory of the triangle, for example, Matgieu’s
theory on conjugate points, on the isogonal similarity of triangles, and area formulas.
8. Bibliography
https://www.youtube.com/watch?app=desktop&v=rsaLcRRlm9c
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1. D. D. Efremov, New geometry of a triangle. - Odessa: type. blank ed. M.
Spencer, 1902. — 334, X, III, III с.
2. 1902. — 334, X, III, III с. — Bibliography in the foreword
3. Karlyuchenko A. V., Filippovsky G. B. The joy of cooperation! - Kyiv, 2016. 127 p., including the region.
4. I. Kushner. Method of isogonal lines Text/I.Kushner//Kvant. -2010.-№6. –S.3536.
5. A.KULIKOVA, D.PROKOPENKO. The exorcism theorem. Text /
A.KULIKOVA, D.PROKOPENKO//Kvant. -2018.-No.4.-P.41-44.
6. Akopyan A. V. Geometry in pictures. - MTSNMO, 2017 - 236s.
7. G.Filippovsky. On the isogonals of a triangle. Text/ G.Filippovsky//Mathematics.
-2021.-No.1.-P.41-43.
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