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). Through the point of intersection of any two bisectors of external angles passes the bisector of an internal angle. 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 inscribed circle 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
.
In fact, BN
=
p
-
b
And AM=AR=SR-AS=p - c.
Ratios rr
c
=(p
- A)(p-V )
And r
p=r
With (p-c) can be used to derive Heron's formula S
2
=
p
(p
-
a
)(p
-
b
)(p
-
c
),
Where S
- area of the 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
HRV 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.
INTERSECTION POINT OF ELEVATIONS
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. Consequently, 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=CP·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. 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 am 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 the 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. This is it 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 consider 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 circumscribed circle 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 a point of the triangle to its vertices will be 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 circumscribed circle 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 heights 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 point $O$ be the intersection point of perpendicular bisectors $n\ and\ m$ (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 middle 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.
In this lesson we will look at four wonderful points of the triangle. Let us dwell on two of them in detail, recall the proofs of important theorems and solve the problem. Let us remember and characterize the remaining two.
Subject:Revision of the 8th grade geometry course
Lesson: Four Wonderful Points of a Triangle
A triangle is, first of all, three segments and three angles, therefore the properties of segments and angles are fundamental.
The segment AB is given. Any segment has a midpoint, and a perpendicular can be drawn through it - let’s denote it as p. Thus, p is the perpendicular bisector.
Theorem (main property of the perpendicular bisector)
Any point lying on the perpendicular bisector is equidistant from the ends of the segment.
Prove that
Proof:
Consider triangles and (see Fig. 1). They are rectangular and equal, because. have a common leg OM, and legs AO and OB are equal by condition, thus, we have two right triangles, equal in two legs. It follows that the hypotenuses of the triangles are also equal, that is, what was required to be proved.
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The converse theorem is true.
Theorem
Each point equidistant from the ends of a segment lies on the perpendicular bisector to this segment.
Given a segment AB, a perpendicular bisector to it p, a point M equidistant from the ends of the segment (see Fig. 2).
Prove that point M lies on the perpendicular bisector of the segment.
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Proof:
Consider a triangle. It is isosceles, as per the condition. Consider the median of a triangle: point O is the middle of the base AB, OM is the median. According to the property of an isosceles triangle, the median drawn to its base is both an altitude and a bisector. It follows that . But line p is also perpendicular to AB. We know that at point O it is possible to draw a single perpendicular to the segment AB, which means that the lines OM and p coincide, it follows that the point M belongs to the straight line p, which is what we needed to prove.
If it is necessary to describe a circle around one segment, this can be done, and there are infinitely many such circles, but the center of each of them will lie on the perpendicular bisector to the segment.
They say that the perpendicular bisector is the locus of points equidistant from the ends of a segment.
A triangle consists of three segments. Let us draw bisectoral perpendiculars to two of them and obtain the point O of their intersection (see Fig. 3).
Point O belongs to the perpendicular bisector to side BC of the triangle, which means it is equidistant from its vertices B and C, let’s denote this distance as R: .
In addition, point O is located on the perpendicular bisector to segment AB, i.e. , at the same time, from here.
Thus, point O of the intersection of two midpoints
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perpendiculars of the triangle is equidistant from its vertices, which means it also lies on the third bisector perpendicular.
We have repeated the proof of an important theorem.
The three perpendicular bisectors of a triangle intersect at one point - the center of the circumcircle.
So, we looked at the first remarkable point of the triangle - the point of intersection of its bisectoral perpendiculars.
Let's move on to the property of an arbitrary angle (see Fig. 4).
The angle is given, its bisector is AL, point M lies on the bisector.
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If point M lies on the bisector of an angle, then it is equidistant from the sides of the angle, that is, the distances from point M to AC and to BC of the sides of the angle are equal.
Proof:
Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, and the angles are equal, since AL is the bisector of the angle. Thus, right triangles are equal in hypotenuse and acute angle, it follows that , which is what needed to be proved. Thus, a point on the bisector of an angle is equidistant from the sides of that angle.
The converse theorem is true.
Theorem
If a point is equidistant from the sides of an undeveloped angle, then it lies on its bisector (see Fig. 5).
An undeveloped angle is given, point M, such that the distance from it to the sides of the angle is the same.
Prove that point M lies on the bisector of the angle.
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Proof:
The distance from a point to a line is the length of the perpendicular. From point M we draw perpendiculars MK to side AB and MR to side AC.
Consider triangles and . These are right triangles and they are equal because... have a common hypotenuse AM, legs MK and MR are equal by condition. Thus, right triangles are equal in hypotenuse and leg. From the equality of triangles follows the equality of the corresponding elements; equal angles lie opposite equal sides, thus, 
Therefore, point M lies on the bisector of the given angle.
If you need to inscribe a circle in an angle, this can be done, and there are infinitely many such circles, but their centers lie on the bisector of a given angle.
They say that a bisector is the locus of points equidistant from the sides of an angle.
A triangle consists of three angles. Let's construct the bisectors of two of them and get the point O of their intersection (see Fig. 6).
Point O lies on the bisector of the angle, which means it is equidistant from its sides AB and BC, let’s denote the distance as r: . Also, point O lies on the bisector of the angle, which means it is equidistant from its sides AC and BC: , , from here.
It is easy to notice that the point of intersection of the bisectors is equidistant from the sides of the third angle, which means it lies on
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angle bisector. Thus, all three bisectors of the triangle intersect at one point.
So, we remembered the proof of another important theorem.
The bisectors of the angles of a triangle intersect at one point - the center of the inscribed circle.
So, we looked at the second remarkable point of the triangle - the point of intersection of the bisectors.
We examined the bisector of an angle and noted its important properties: the points of the bisector are equidistant from the sides of the angle, in addition, the tangent segments drawn to the circle from one point are equal.
Let us introduce some notation (see Fig. 7).
Let us denote equal tangent segments by x, y and z. The side BC lying opposite the vertex A is designated as a, similarly AC as b, AB as c.
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Problem 1: in a triangle, the semi-perimeter and length of side a are known. Find the length of the tangent drawn from the vertex A - AK, denoted by x.
Obviously, the triangle is not completely defined, and there are many such triangles, but it turns out that they have some elements in common.
For problems involving an inscribed circle, the following solution method can be proposed:
1. Draw bisectors and get the center of the inscribed circle.
2. From center O, draw perpendiculars to the sides and obtain points of tangency.
3. Mark equal tangents.
4. Write out the relationship between the sides of the triangle and the tangents.
© Kugusheva Natalya Lvovna, 2009 Geometry, 8th grade TRIANGLE FOUR REMARKABLE POINTS
The intersection point of the medians of a triangle The intersection point of the bisectors of a triangle The intersection point of the altitudes of a triangle The intersection point of the perpendicular bisectors of a triangle
The median (BD) of a triangle is the segment that connects the vertex of the triangle to the midpoint of the opposite side. A B C D Median
The medians of a triangle intersect at one point (the center of gravity of the triangle) and are divided by this point in a ratio of 2: 1, counting from the vertex. AM: MA 1 = VM: MV 1 = SM:MS 1 = 2:1. A A 1 B B 1 M C C 1
The bisector (A D) of a triangle is the bisector segment of the interior angle of the triangle.
Each point of the bisector of an undeveloped angle is equidistant from its sides. Conversely: every point lying inside an angle and equidistant from the sides of the angle lies on its bisector. A M B C
All bisectors of a triangle intersect at one point - the center of the circle inscribed in the triangle. C B 1 M A V A 1 C 1 O The radius of a circle (OM) is a perpendicular dropped from the center (TO) to the side of the triangle
HEIGHT The altitude (C D) of a triangle is the perpendicular segment drawn from the vertex of the triangle to the straight line containing the opposite side. A B C D
The altitudes of a triangle (or their extensions) intersect at one point. A A 1 B B 1 C C 1
MIDPERPENDICULAR The perpendicular bisector (DF) is the line perpendicular to the side of the triangle and dividing it in half. A D F B C
A M B m O Each point of the perpendicular bisector (m) to a segment is equidistant from the ends of this segment. Conversely: every point equidistant from the ends of a segment lies on the perpendicular bisector to it.
All perpendicular bisectors of the sides of a triangle intersect at one point - the center of the circle circumscribed about the triangle. A B C O The radius of the circumscribed circle is the distance from the center of the circle to any vertex of the triangle (OA). m n p
Tasks for students Construct a circle inscribed in an obtuse triangle using a compass and ruler. To do this: Construct bisectors in an obtuse triangle using a compass and ruler. The point of intersection of the bisectors is the center of the circle. Construct the radius of the circle: a perpendicular from the center of the circle to the side of the triangle. Construct a circle inscribed in the triangle.
2. Using a compass and ruler, construct a circle circumscribing an obtuse triangle. To do this: Construct perpendicular bisectors to the sides of the obtuse triangle. The point of intersection of these perpendiculars is the center of the circumscribed circle. The radius of a circle is the distance from the center to any vertex of the triangle. Construct a circle around the triangle.
