γ+(90° +γ/2) =180°, which means γ = 60°. In this case the chords O.A.
1
And OB 1
circumcircle of a quadrilateral OA 1
NE 1
are equal because they have equal angles OCA
1
And SALT 1
.
Inscribed circle Δ ABC touches its sides at internal points. Let's find out what kind of circles there are that touch three lines AB, BC And SA. The center of a circle tangent to two intersecting lines lies on one of the two lines bisecting the angles between the original lines. Therefore, the centers of circles tangent to straight lines AB, BC And S A, lie on the bisectors of the external or internal angles of the triangle (or their extensions). The bisector of an interior angle passes through the intersection point of any two exterior angle bisectors. The proof of this statement repeats verbatim the proof of the corresponding statement for the bisectors of interior angles. As a result, we get 4 circles with centers O, ABOUT A , Oh And ABOUT With
(Fig. 57). Circle with center ABOUT A
touches the side Sun And
continuations of the parties AB And AC; this circle is called uninscribed
circumference Δ ABC. The radius of the incircle of a triangle is usually denoted by r, and the radii of excircles by r A ,
G b and g With .
The following relations hold between the radii of the inscribed and excircle circles:
G /
g s =(р-с)/р and G
G With =(p - a) (p - b), Where R- semi-perimeter Δ ABC. Let's prove it. Let K and L be the points of tangency of the inscribed and excircle with the line Sun(Fig. 58). Right Triangles JUICE And CO
c
L
are similar, therefore
G /
g s =OK/O With L
=
CK
/
C.L.
..
It was previously proven that SC = (a+b-c)/2=p-c.
It remains to check that C.L.
=
p
.
Let M And R- points of tangency of an excircle with straight lines AB And AC. Then
CL= (CL+CP)/ 2 = (CB+BL+CA+AP)/2 = (CB+BM + CA+AM)/2 = R
To prove the relation rr
c
=(p
-
a
)(p
-
b
)
consider right triangles L.O.
C
B
And KVO, which are similar because
<OBK
+<
O
C
B.L.
=(<СВА + <АВ
L
)/2=90°.
Means, L O s /ВL =BK /KO, i.e. rr
c
=
K.O.
·
L.O.
c
=
B.K.
·
B.L.
.
It remains to note that VK=(a
+
c
-
b
)/2=
p
-
b
And B.L.
=
C.L.
-
C.B.
=
p
-
a
.
Let us note one more interesting property (already actually proven along the way). Let the inscribed and excircle touch the side AB at points N And M(Fig. 58). Then A.M.
=
BN
.
Indeed, BN
=
p
-
b
And AM=AR=SR-AS=p - c.
Ratios rr
c
=(p
- A)(p-V )
And r
p=r
With (R-c) can be used to derive Heron's formula S
2
=
p
(p
-
a
)(p
-
b
)(p
-
c
),
Where S
- area of a triangle. Multiplying these ratios, we get r
2
p
=(p
-
a
)(p
-
b
)(p
-
c
).
It remains to check that S
=
pr
.
This can be easily done by cutting Δ ABC on ΔAOB,
ΔBOS And ΔSOA.
MEDIAN INTERSECTION POINT
Let us prove that the medians of a triangle intersect at one point. For this, consider the point M, where the medians intersect AA 1
And BB 1
.
Let's carry out in Δ BB1S midline A
1
A
2
,
parallel BB 1
(Fig. 59). Then A
1
M
:
A.M.
=
B
1
A
2
:
AB
1
=
B
1
A
2
:
B
1
C
=
B.A.
1
:VS=1:2, i.e. the point of intersection of the medians BB 1
And AA 1
divides the median AA 1
in a ratio of 1:2. Similarly, the intersection point of the medians SS 1
And AA 1
divides the median AA 1
in a ratio of 1:2. Therefore, the intersection point of the medians AA 1
And BB 1
coincides with the intersection point of the medians AA 1
And SS 1
.
If the intersection point of the medians of a triangle is connected to the vertices, then the triangle will be divided into three triangles of equal area. Indeed, it is enough to prove that if R- any point of the median AA 1
V ABC, then the area ΔAVR And ΔACP are equal. After all, medians AA 1
And RA 1
in Δ ABC and Δ RVS cut them into triangles of equal area.
The converse statement is also true: if for some point R, lying inside Δ ABC, area Δ AVR, Δ
ON WEDNESDAY And ΔSAR are equal, then R- point of intersection of medians. In fact, from the equality of areas ΔAVR And ΔHRV it follows that the distances from points A and C to the straight line VR are equal, which means VR passes through the middle of the segment AC. For AR And SR the proof is similar.
The equality of the areas of the triangles into which the medians divide the triangle allows us to find the ratio of the area s of a triangle composed of medians as follows ΔABC, to the area S of Δ itself ABC. Let M- point of intersection of medians Δ ABC; dot A" symmetrical A relative to the point M(Fig. 60)
On the one hand, the area ΔA"MS equal to S/3. On the other hand, this triangle is composed of segments, the length of each of which is equal to 2/3 of the length of the corresponding median, so its area
equal to (2/3) 2 s = 4s /9. Hence, s
=3
S
/4.
A very important property of the intersection point of the medians is that the sum of the three vectors going from it to the vertices of the triangle is equal to zero. Let us first note that AM=1/3(AB+AC), Where M- point of intersection of medians Δ
ABC
.
In fact, if
ABA
"WITH- parallelogram, then AA"=AB+AC And AM=1/3AA". That's why MA+MV+MC=1/3(BA+SA+AB + SV + AC + BC) = 0.
It is also clear that only the point of intersection of medians has this property, since if X
- any other point, then
HA+XB+XC=(XM+MA)+(XM+MV)+(XM+MS)=3ХМ..
Using this property of the point of intersection of the medians of a triangle, we can prove the following statement: the point of intersection of the medians of a triangle with the vertices at the midpoints of the sides AB,CD
And E.F.
hexagon ABCDEF
coincides with the point of intersection of the medians of the triangle with the vertices at the midpoints of the sides sun,DE
And F.A.
.
In fact, taking advantage of the fact that if, for example, R- the middle of the segment AB, then for any point X
equality is true HA+ HB=2ХР, it is easy to prove that the intersection points of the medians of both triangles under consideration have the property that the sum of the vectors going from them to the vertices of the hexagon is equal to zero. Therefore, these points coincide.
The intersection point of the medians has one property that sharply distinguishes it from the other remarkable points of the triangle: if Δ A"B"C" is a projection ΔABC onto the plane, then the point of intersection of the medians Δ A "B" C"
is the projection of the intersection point of the medians ΔABC on the same plane. This easily follows from the fact that when projecting, the middle of the segment goes into the middle of its projection, which means that the median of the triangle goes into the median of its projection. Neither the bisector nor the height have this property.
It should be noted that the point of intersection of the medians of a triangle is its center of mass, both the center of mass of a system of three material points with equal masses located at the vertices of the triangle, and the center of mass of a plate shaped like a given triangle. The equilibrium position of a triangle hinged at an arbitrary point X
,
there will be a position in which the beam HM directed towards the center of the Earth. For a triangle hinged at the intersection point of the medians, any position is an equilibrium position. In addition, a triangle whose median intersection point rests on the needle tip will also be in an equilibrium position.
ELEVATION INTERSECTION POINT
To prove that the heights Δ ABC intersect at one point, recall the path of proof outlined at the end of the section “Center of the Circumscribed Circle”. Let's take you through the peaks A, B And WITH straight lines parallel to opposite sides; these lines form Δ A 1
IN 1
WITH 1
(Fig. 61). Heights Δ ABC are the perpendicular bisectors to the sides ΔA
1
B
1
C
1
.
Consequently, they intersect at one point - the center of the circumcircle ΔA
1
B
1
C
1
.
The point of intersection of the altitudes of a triangle is sometimes called its orthocenter.
-
It is easy to check that if H is the point of intersection of heights Δ ABC, That A, B And WITH - height intersection points Δ VNS, ΔSNA and Δ ANV respectively.
It is also clear that<ABC
+ <
A.H.C.
=
180° because <
B.A.
1
H
= <
B.C.
1
H
=90° (A
1
And C
1
- bases of heights). If the point H
1
symmetrical to point H relative to the straight line AC, then a quadrilateral ABCN 1
inscribed. Therefore, the radii of circumscribed circles Δ ABC and Δ AN S are equal and these circles are symmetrical with respect to the side AC(Fig. 62). Now it is easy to prove that
AN=a|ctg A|, where a=BC. Indeed,
AH=2R sin<
ACH=2R|cos A| =a|ctg A| .
Let's assume for simplicity that ΔABC acute-angled and consider Δ A
1
B
1
C
1
,
formed by the bases of its heights. It turns out that the center of the inscribed circle Δ A
1
B
1
C
1
is the point of intersection of the heights Δ ABC, and the centers of excircles
ΔA
1
B
1
C
1
are the vertices of Δ ABC(Fig. 63). Points A 1
And IN 1
CH(since the corners NV 1
S and ON 1
WITH straight), so <
H.A.
1
B
1
= <
HCB
1
.
Likewise<H.A.
1
C
1
= <
HBC
1
.
And since<HCB
1
= =<
HBC
1
That A 1
A - bisector<IN 1
A 1
WITH 1
.
Let N- point of intersection of heights AA 1
, BB 1
And CC
1
triangle ABC
.
Points A
1
And IN 1
lie on a circle with diameter AB, That's why A.H.
·
A
1
H
=
B.H.
·
B
1
H
. Likewise VNB
1
H
=CH ·C 1
N.
For an acute triangle, the converse statement is also true: if points A 1, B
1
And C
1
lie on the sides VS, SA and AB acute-angled Δ ABC and segments AA 1
, BB 1
And SS 1
intersect at a point R, and AR A 1
Р=ВР·В 1
P=SR·S 1
R, That R- point of intersection of heights. In fact, from the equality
AP ·A 1 P =BP ·B 1 P
it follows that the points A, B, A 1
And IN 1
lie on the same circle with the diameter AB, which means <
AB
1
B
= <
B.A.
1
A
=γ.
Likewise <
ACiC
=<
CAiA
=
β
And
<СВ
1
B=<ВС
1
C= α
(Fig. 64). It is also clear that α + β= CC
1
A
=
l
80°, β+γ=180° and γ + α = 180°. Therefore, α = β=γ=90°.
The point of intersection of the altitudes of a triangle can be determined in another very interesting way, but for this we need the concepts of a vector and a scalar product of vectors.
Let ABOUT- center of the circumcircle Δ ABC. Vector sum O A+ O.B.
+ OS is some vector, so there is such a point R, What OR = OA + OB+OS. It turns out that R- point of intersection of heights Δ ABC!
Let us prove, for example, that AP
perpendicular B.C.
.
It's clear that AR=AO+
+op=ao+(oa+ov+os)=ov+os and all= -ov+os.
Therefore, the scalar product of vectors AR And Sun equals OS 2
- O.B.
2
=
R
2
-
R
2
=0,
i.e. these vectors are perpendicular.
This property of the orthocenter of a triangle allows us to prove some far from obvious statements. Consider, for example, a quadrilateral ABCD
,
inscribed in a circle. Let Na, Nv, Ns And H
d
- orthocenters Δ
BCD
, Δ
CDA
, Δ
DAB
and Δ
ABC
respectively. Then the midpoints of the segments AN A , VN, CH WITH ,
D.H.
d
match up. In fact, if ABOUT is the center of the circle, and M- the middle of the segment AN A ,
That OM=1/2(0A + OH A )= =1/2(OA + OB+OS+OD
)
.
For the midpoints of the other three segments we obtain exactly the same expressions.
EULER DIRECT
The most amazing property of the wonderful dots isthe angle is that some of them are connected to each otherby certain ratios. For example, the intersection point median M,
the point of intersection of the heights H and the center of the circumscribed circleproperties O lie on the same straight line, and the pointM divides the segment HE
so that the relation is validOM:MN= 1:2. This the theorem was proven in 1765 by Leonhard Euler, whoWith his tireless activity, he significantly developed many areas of mathematics and laid the foundations for many of its new branches. He was born in 1707 in Switzerland. At age 20, Euler recommendedBernoulli brothers received an invitation to come to St. Petersburgburg, where an academy was organized shortly before. INat the end of 1740 in Russia in connection with the rise to power of Anna LeopolDovna, an alarming situation developed, and Euler moved toBerlin. After 25 years, he returned to Russia again, in totalEuler lived in St. Petersburg for more than 30 years. While in Burleyno, Euler maintained close contact with the Russian Academy and wasits honorary member. From Berlin Euler corresponded with Lomonoowls Their correspondence began as follows. In 1747, Lomonosov was elected a professor, that is, a full member of the academy; The empress approved this election. After thatreactionary Academy official Schumacher, who vehemently hates LawMonosov, sent his work to Euler, hoping to get information about thembad review. (Euler was only 4 years older than Lomonosov,but his scientific authority was already very high by that time.)In his review, Euler wrote: “All these works are not only goodshi, but also excellent, because he explains physical and chemical the most necessary and difficult matters, which are completely unknown and interpretations were impossibleto the most witty and learnedfamous people, with such a founderthing that I'm quite sure aboutthe accuracy of his evidence...One must wish that everythingwhich academies were able to show such inventions thatwhich Mr. Lomo showed noses."
Let's move on to the proof Euler's theorem. Let's consider Δ
A
1
B
1
C
1
with vertices in 
midpoints of the sides Δ ABC; let H
1
and H - their orthocenters (Fig. 65). Point H 1 coincides with the center ABOUT circumcircle Δ ABC. Let us prove that Δ C
1
H
1
M
=Δ
CHM
.
Indeed, by the property of the intersection point of medians WITH 1
M:
CM= 1:2, similarity coefficient Δ A
1
B
1
C
1
and Δ ABC is equal to 2, so C
1
H
1
:
CH
=1:2,
Besides,<H
1
C
1
M
=<НСМ (C
1
H
1
||
CH
).
Therefore,< C
1
M.H.
1
= < SMN, which means point M lies on the segment H
1
H
.
Besides, H
1
M
:
M.H.
=1:2,
since the similarity coefficient Δ C
1
H
1
M
and Δ SNM equals 2.
CIRCLE OF NINE POINTS
In 1765, Euler discovered that the midpoints of the sides of a triangle and the bases of its altitudes lie on the same circle. We will also prove this property of a triangle.
Let B 2 be the base of the height dropped from the top IN on
side AC. Points IN and B 2 are symmetrical about a straight line A 1
WITH 1
(Fig. 66). Therefore, Δ A
1
IN
2
WITH
1
= Δ
A
1
B.C.
t
= Δ
A
1
B
1
C
1
,
That's why <
A
1
B
2
C
1
= <А
1
IN 1
WITH 1
,
which means point IN 2
lies on the described
circle ΔA
1
IN
1
WITH
1
.
For the remaining bases of heights the proof is similar. „
Subsequently, it was discovered that three more points lie on the same circle - the midpoints of the segments connecting the orthocenter with the vertices of the triangle. That's what it is circle of nine points.
Let Az And NW- midpoints of segments AN And CH, S 2
- the base of the height dropped from the top WITH on AB(Fig. 67). Let us first prove that A
1
C
1
A
3
C
3
- rectangle. This easily follows from the fact that A 1
NW And A
3
C
1
- midlines Δ VSN And ΔAVN, A A
1
C
1
And A 3
NW- midlines Δ ABC and Δ ASN. Therefore the points A 1
And Az lie on a circle with diameter WITH 1
NW, and since Az And NW lie on a circle passing through the points A 1,
C
1
and C 2. This circle coincides with the circle considered by Euler (if Δ ABC not isosceles). For a point Vz the proof is similar.
TORRICELLI POINT
Inside an arbitrary quadrilateral ABCD
It is easy to find the point whose sum of distances to the vertices has the smallest value. Such a point is a point ABOUT intersection of its diagonals. In fact, if X
- any other point, then AH+HS≥AC=AO+OS And BX
+
XD
≥
BD
=
B.O.
+
O.D.
,
and at least one of the inequalities is strict. For a triangle, a similar problem is more difficult to solve; we will now move on to solving it. For simplicity, we will analyze the case of an acute triangle.
Let M- some point inside the acute-angled Δ ABC. Let's turn it around Δ ABC along with the dot M 60° around the point A(Fig. 68). (More precisely, let B, C And M"- images of points B, C And M when rotated 60° around a point A.) Then AM+VM+SM=MM"+B.M.
+
C
"
M
", AM=MM", So as ΔAMM"- isosceles (AM=AM") And<MAM" = 60°. The right side of the equality is the length of the broken line VMM"S"
;
it will be smallest when this broken line
coincides with the segment Sun"
.
In this case<.
A.M.B.
=
180° -<AMM" = 120° and<АМС = <A.M.
"
C
-
180°-<A.M.
"
M
=
120°, i.e. sides AB, BC and SA are visible from the point M at an angle of 120°. Such a point M called Torricelli point triangle ABC
.
Let us prove, however, that inside an acute triangle there always exists a point M, from which each side is visible at an angle of 120°. Let's build it on the side AB triangle ABC
externally correct Δ ABC 1
(Fig. 69). Let M-point of intersection of the circumcircle ΔABC 1
and straight SS 1
.
Then ABC
1
=60° And ABC visible from the point M at an angle of 120°. Continuing these arguments a little further, we can obtain another definition of the Torricelli point. Let's build regular triangles A 1
Sun And AB 1
WITH also on the sides of the Armed Forces and AC. Let us prove that point M also lies on the line AA 1
.
Indeed, period M lies on the circumcircle Δ A
1
B.C.
,
That's why<A
1
M.B.
= <
A
1
C.B.
= 60°, which means<A 1
MV+<.
B.M.A.
=
180°. Likewise point M lies on a straight line BB 1
(Fig. 69).
Inside Δ ABC there is a single point M from which its sides are visible at an angle of 120°, because the circumscribed circles Δ ABC
1
, Δ
AB
i
C
and Δ A
1
Sun cannot have more than one common point.
Let us now give a physical (mechanical) interpretation of the Torricelli point. Let us fix Δ at the vertices ABC rings, we pass three ropes through them, one ends of which are tied, and loads of equal mass are attached to the other ends (Fig. 70). If x = MA, y = MV,z
=
M.C.
And A is the length of each thread, then the potential energy of the system under consideration is equal to m g
(x
-A)+ m g
(y
-
a
)+
mg
(z
--A). At the equilibrium position, the potential energy has the smallest value, so the sum x+y+z also has the smallest value. On the other hand, in the equilibrium position the resultant of forces at the point M equal to zero. These forces are equal in absolute magnitude, therefore the pairwise angles between the force vectors are equal to 120°.
It remains to tell how things stand in the case of an obtuse triangle. If the obtuse angle is less than 120°, then all previous arguments remain valid. And if the obtuse angle is greater than or equal to 120°, then the sum of the distances from the point of the triangle to its vertices will be the smallest when this point is the vertex of the obtuse angle.
BROKARD'S POINTS
Brocard points Δ ABC such internal points are called R And Q
,
What<ABP
= <.
BCP
=<
CAP
And<.
QAB
= <.
QBC
= < QCA
(for an equilateral triangle, the Brocard points merge into one point). Let us prove that inside any Δ ABC there is a point R, having the required property (for a point Q
the proof is similar). Let us first formulate the definition of the Brocard point in a different form. Let us denote the angle values as shown in Figure 71. Since<ARV=180° - a+x-y, equality x=y is equivalent to equality<APB
=180°-<
.
A
.
Hence, R- point Δ ABC, from which sides AB,
Sun And SA visible at angles of 180° -<.
A
,
180°-<B
,
180°-<WITH.
Such a point can be constructed as follows. Let's build on
side Sun triangle ABC similar triangle CA1B
as shown in Figure 72. Let us prove that the point P of intersection of the straight line AA1 and circumcircle ΔA1BC sought after. In fact,<BPC
=18
O
° - β
And<APB
=
180°-<A
t
P.B.
=
180° -<A
1
C.B.
=
l
80°- A. Let us further construct similar triangles on the sides in a similar way AC And AB(Fig. 73). Because<.
APB
=
180° - A, dot R also lies on the circumcircle Δ ABC 1
Hence,<BPC
1
= <BAC
1
= β, which means point
R lies on the segment SS 1
.
It lies similarly on the segment BB 1
,
i.e. R - point of intersection of segments AA 1
, BB 1
And SS 1
.
Brocard's point R has the following interesting property. Let straight AR, VR And SR intersect the circumcircle ΔABC
at points A 1, B 1 and C 1 (Fig. 74). Then Δ ABC = Δ
B
1
WITH 1
A
1
.IN in fact,<.
A
1
B
1
C
1
= <
A
1
B
1
B
+ <
BB 1 C 1 =<A
1
AB
+<В
CC 1 =<A
1
AB
+ +<
A
1
A.C.
=<.ВАС,
by the property of the Brocard point ΔABC, the angles BCC 1 and A 1 AC are equal, which means A
1
C
1
=
B.C.
.
Equality of the remaining sides Δ ABC and Δ B 1 C 1 A 1 are checked in the same way.
In all the cases we have considered, the proof that the corresponding triples of lines intersect at one point can be carried out using Ceva's theorem. We will formulate this theorem.
Theorem. Let on the sides AB, BC And S A triangle ABC
points taken WITH 1
, A 1
And IN 1
respectively. Direct AA 1
, BB 1
And SS 1
intersect at one point if and only if
AC 1 / C 1 V VA 1 / A 1 C SV 1 / V 1 A = 1.
The proof of the theorem is given in the textbook on geometry for grades 7-9 by L.S. Atanasyan on p. 300.
Literature.
1.Atanasyan L.S. Geometry 7-9.- M.: Education, 2000.
2. Kiselev A.P. Elementary geometry. - M.: Education, 1980.
3. Nikolskaya I.L. Optional course in mathematics. M.: Education, 1991.
4. Encyclopedic Dictionary of a Young Mathematician.. Comp. A.P.Savin.-.M.: Pedagogy, 1989.
There are so-called four remarkable points in a triangle: the point of intersection of the medians. The point of intersection of bisectors, the point of intersection of altitudes and the point of intersection of perpendicular bisectors. Let's look at each of them.
Intersection point of triangle medians
Theorem 1
On the intersection of medians of a triangle: The medians of a triangle intersect at one point and are divided by the intersection point in the ratio $2:1$ starting from the vertex.
Proof.
Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ are its medians. Since medians divide the sides in half. Let's consider the middle line $A_1B_1$ (Fig. 1).
Figure 1. Medians of a triangle
By Theorem 1, $AB||A_1B_1$ and $AB=2A_1B_1$, therefore, $\angle ABB_1=\angle BB_1A_1,\ \angle BAA_1=\angle AA_1B_1$. This means that triangles $ABM$ and $A_1B_1M$ are similar according to the first criterion of similarity of triangles. Then
Similarly, it is proved that
The theorem has been proven.
Intersection point of triangle bisectors
Theorem 2
On the intersection of bisectors of a triangle: The bisectors of a triangle intersect at one point.
Proof.
Consider triangle $ABC$, where $AM,\BP,\CK$ are its bisectors. Let the point $O$ be the intersection point of the bisectors $AM\ and\BP$. Let us draw perpendiculars from this point to the sides of the triangle (Fig. 2).
Figure 2. Triangle bisectors
Theorem 3
Each point of the bisector of an undeveloped angle is equidistant from its sides.
By Theorem 3, we have: $OX=OZ,\ OX=OY$. Therefore, $OY=OZ$. This means that the point $O$ is equidistant from the sides of the angle $ACB$ and, therefore, lies on its bisector $CK$.
The theorem has been proven.
The point of intersection of the perpendicular bisectors of a triangle
Theorem 4
The perpendicular bisectors to the sides of a triangle intersect at one point.
Proof.
Let a triangle $ABC$ be given, $n,\ m,\ p$ its perpendicular bisectors. Let the point $O$ be the intersection point of the bisectoral perpendiculars $n\ and\ m$ (Fig. 3).
Figure 3. Perpendicular bisectors of a triangle
To prove it, we need the following theorem.
Theorem 5
Each point of the perpendicular bisector to a segment is equidistant from the ends of the segment.
By Theorem 3, we have: $OB=OC,\ OB=OA$. Therefore, $OA=OC$. This means that the point $O$ is equidistant from the ends of the segment $AC$ and, therefore, lies on its perpendicular bisector $p$.
The theorem has been proven.
Point of intersection of triangle altitudes
Theorem 6
The altitudes of a triangle or their extensions intersect at one point.
Proof.
Consider triangle $ABC$, where $(AA)_1,\ (BB)_1,\ (CC)_1$ is its altitude. Let us draw a straight line through each vertex of the triangle parallel to the side opposite the vertex. We get a new triangle $A_2B_2C_2$ (Fig. 4).
Figure 4. Triangle heights
Since $AC_2BC$ and $B_2ABC$ are parallelograms with a common side, then $AC_2=AB_2$, that is, point $A$ is the midpoint of side $C_2B_2$. Similarly, we find that point $B$ is the midpoint of side $C_2A_2$, and point $C$ is the midpoint of side $A_2B_2$. From the construction we have that $(CC)_1\bot A_2B_2,\ (BB)_1\bot A_2C_2,\ (AA)_1\bot C_2B_2$. Therefore, $(AA)_1,\ (BB)_1,\ (CC)_1$ are the perpendicular bisectors of triangle $A_2B_2C_2$. Then, by Theorem 4, we have that the heights $(AA)_1,\ (BB)_1,\ (CC)_1$ intersect at one point.
Baranova Elena
This work examines the remarkable points of the triangle, their properties and patterns, such as the nine-point circle and the Euler straight line. The historical background of the discovery of Euler's straight line and the nine-point circle is given. The practical direction of application of my project is proposed.
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"WONDERFUL POINTS OF A TRIANGLE." (Applied and fundamental questions of mathematics) Elena Baranova 8th grade, MKOU “Secondary School No. 20” Pos. Novoizobilny, Dukhanina Tatyana Vasilievna, mathematics teacher, Municipal Educational Institution "Secondary School No. 20" Novoizobilny village 2013. Municipal government educational institution "Secondary school No. 20"
Goal: study the triangle for its remarkable points, study their classifications and properties. Objectives: 1. Study the necessary literature 2. Study the classification of remarkable points of a triangle 3.. Get acquainted with the properties of remarkable points of a triangle 4. Be able to construct remarkable points of a triangle. 5. Explore the scope of remarkable points. Object of study - section of mathematics - geometry Subject of study - triangle Relevance: expand your knowledge about the triangle, the properties of its remarkable points. Hypothesis: connection between triangle and nature
The point of intersection of the perpendicular bisectors. It is equidistant from the vertices of the triangle and is the center of the circumscribed circle. Circles circumscribed about triangles, the vertices of which are the midpoints of the sides of the triangle and the vertices of the triangle intersect at one point, which coincides with the point of intersection of the perpendicular bisectors.
Intersection point of bisectors The intersection point of bisectors of a triangle is equidistant from the sides of the triangle. OM=OA=OB
Point of intersection of altitudes The point of intersection of the bisectors of a triangle, the vertices of which are the bases of the heights, coincides with the point of intersection of the altitudes of the triangle.
Intersection point of medians The medians of a triangle intersect at one point, which divides each median in a ratio of 2:1, counting from the vertex. If the intersection point of the medians is connected to the vertices, then the triangle will be divided into three triangles of equal area. An important property of the intersection point of the medians is the fact that the sum of vectors, the beginning of which is the intersection point of the medians, and the ends are the vertices of the triangles, is equal to zero M1 N C B A m2 m3 M1 N C B A m2 m3 M1 N C B A m2 m3 M1 N C B A m2 m3
Torricelli point Note: A Torricelli point exists if all angles of the triangle are less than 120.
Circle of nine points B1, A1, C1 – bases of heights; A2, B2, C2 – the midpoints of the corresponding sides; A3, B3, C3, are the midpoints of segments AN, VN and CH.
Euler's straight line The point of intersection of medians, the point of intersection of heights, the center of a circle of nine points lie on one straight line, which is called Euler's straight line in honor of the mathematician who determined this pattern.
A little from the history of the discovery of remarkable points In 1765, Euler discovered that the midpoints of the sides of a triangle and the bases of its altitudes lie on the same circle. The most amazing property of the remarkable points of a triangle is that some of them are connected to each other by a certain ratio. The intersection point of the medians M, the intersection point of the heights H, and the center of the circumcircle O lie on the same straight line, and the point M divides the segment OH so that the relation OM: OH = 1: 2 is valid. This theorem was proven by Leonhard Euler in 1765.
The connection between geometry and nature. In this position, the potential energy has the smallest value and the sum of the segments MA+MB+MC will be the smallest, and the sum of the vectors lying on these segments with the beginning at the Torricelli point will be equal to zero.
Conclusions I learned that in addition to the wonderful points of intersection of heights, medians, bisectors and perpendicular bisectors that I know, there are also wonderful points and lines of a triangle. I will be able to use the knowledge gained on this topic in my educational activities, independently apply theorems to certain problems, and apply the learned theorems in a real situation. I believe that using the wonderful points and lines of a triangle in learning mathematics is effective. Knowing them significantly speeds up the solution of many tasks. The proposed material can be used both in mathematics lessons and in extracurricular activities for students in grades 5-9.
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Silchenkov Ilya
materials for the lesson, presentation with animation
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The midline of a triangle is a segment connecting the midpoints of its two sides and is equal to half of that side. Also, according to the theorem, the middle line of a triangle is parallel to one of its sides and is equal to half of this side.
If a line is perpendicular to one of two parallel lines, then it is also perpendicular to the other
Remarkable points of the triangle
Remarkable points of a triangle Intersection point of medians (centroid of triangle); The point of intersection of the bisectors, the center of the inscribed circle; The point of intersection of the perpendicular bisectors; Point of intersection of heights (orthocenter); Euler's straight line and nine-point circle; Gergonne and Nagel points; Point Fermat-Torricelli;
Median intersection point
The median of a triangle is a segment connecting the vertex of any angle of the triangle with the middle of the opposite side.
I. The medians of a triangle intersect at one point, which divides each median in a ratio of 2:1, counting from the vertex.
Proof:
A B C A 1 C 1 B 1 1 2 3 4 0 1. Let us denote by the letter O the point of intersection of two medians AA 1 and B B1 of triangle ABC and draw the middle line A 1 B 1 of this triangle. 2. The segment A 1 B 1 is parallel to side AB and 1/2 AB = A 1 B 1 i.e. AB = 2A1B1 (according to the theorem about the midline of the triangle), therefore 1 = 4 and 3 = 2 (since they internal crosswise angles with parallel lines AB and A 1 B 1 and secant BB 1 for 1, 4 and AA 1 for 3, 2 3. Consequently, triangles AOB and A 1 OB 1 are similar in two angles, and, therefore, their sides are proportional , i.e. the ratios of the sides AO and A 1 O, BO and B 1 O, AB and A 1 B 1 are equal. But AB = 2A 1 B 1, therefore AO = 2A 1 O and BO = 2B 1 O. Thus , the point O of the intersection of the medians BB 1 and AA 1 divides each of them in the ratio 2:1, counting from the vertex. The theorem can be proven similarly for the other two medians.
The center of mass is sometimes called the centroid. That is why they say that the point of intersection of the medians is the centroid of the triangle. The center of mass of a homogeneous triangular plate is located at the same point. If such a plate is placed on a pin so that the tip of the pin hits exactly the centroid of the triangle, then the plate will be in equilibrium. Also, the point of intersection of the medians is the center of the inscribed circle of its medial triangle. An interesting property of the point of intersection of medians is associated with the physical concept of the center of mass. It turns out that if you place equal masses at the vertices of a triangle, then their center will fall exactly at this point.
Bisector intersection point
The bisector of a triangle is a segment of the angle bisector connecting the vertex of one of the angles of the triangle with a point lying on the opposite side.
The bisectors of a triangle intersect at one point equidistant from its sides.
Proof:
C A B A 1 B 1 C 1 0 1. Let us denote by the letter O the point of intersection of bisectors AA 1 and BB 1 of triangle ABC. 3. Let us take advantage of the fact that each point of the bisector of an undeveloped angle is equidistant from its sides and vice versa: each point lying inside the angle and equidistant from the sides of the angle lies on its bisector. Then OK=OL and OK=OM. This means OM=OL, i.e. point O is equidistant from the sides of triangle ABC and, therefore, lies on the bisector CC1 of angle C. 4. Consequently, all three bisectors of triangle ABC intersect at point O. K L M The theorem is proven. 2.Draw from this point the perpendiculars OK, OL and OM, respectively, to the straight lines AB, BC and CA.
Intersection point of perpendicular bisectors
A perpendicular bisector is a line passing through the middle of a given segment and perpendicular to it.
The perpendicular bisectors to the sides of a triangle intersect at one point equidistant from the vertices of the triangle.
Proof:
B C A m n 1. Let us denote by the letter O the point of intersection of the bisectoral perpendiculars m and n to sides AB and BC of triangle ABC. O 2. Using the theorem that each point of the perpendicular bisector to a segment is equidistant from the ends of this segment and vice versa: each point equidistant from the ends of the segment lies on the perpendicular bisector to it, we obtain that OB = OA and OB = OC. 3. Therefore OA = OC, i.e. point O is equidistant from the ends of the segment AC and, therefore, lies on the perpendicular bisector to this segment. 4. Consequently, all three bisectors m, n and p to the sides of triangle ABC intersect at point O. The theorem is proven. R
Point of intersection of heights (or their extensions)
The altitude of a triangle is the perpendicular drawn from the vertex of any angle of the triangle to the straight line containing the opposite side.
The altitudes of a triangle or their extensions intersect at one point, which may lie within the triangle or may be outside it.
Proof:
Let us prove that the lines AA 1, BB 1 and CC 1 intersect at one point. B A C C2 C1 A1 A2 B 1 B 2 1. Draw through each vertex of triangle ABC a straight line parallel to the opposite side. We get triangle A 2 B 2 C 2. 2. Points A, B and C are the midpoints of the sides of this triangle. Indeed, AB=A 2 C and AB=CB 2 are like opposite sides of parallelograms ABA 2 C and ABCB 2, therefore A 2 C=CB 2. Similarly, C 2 A=AB 2 and C 2 B=BA 2. In addition, as follows from the construction, CC 1 is perpendicular to A 2 B 2, AA 1 is perpendicular to B 2 C 2 and BB 1 is perpendicular to A 2 C 2 (from the corollary to the theorem of parallel lines and secants). Thus, lines AA 1, BB 1 and CC 1 are the perpendicular bisectors to the sides of the triangle A 2 B 2 C 2. Therefore, they intersect at one point. The theorem has been proven.
