1 Additional construction leading to the triangle midline theorem, trapezoid and similarity properties of triangles.
And she equal to half the hypotenuse.
Corollary 1.
Corollary 2.
From this it is clear that 
1 All right triangles with the same acute angle are similar. A look at trigonometric functions.
Triangles with hatched and non-hatched sides are similar in that their two angles are equal. Therefore where
This means that these relationships depend only on acute angle right triangle and essentially define it. This is one of the reasons for the appearance trigonometric functions:
Often writing trigonometric functions of angles in similar right triangles is clearer than writing similarity relations!
2 An example of an additional construction is a height lowered to the hypotenuse. Derivation of the Pythagorean theorem based on the similarity of triangles.
Let us lower the height CH to the hypotenuse AB. We have three similar triangles ABC, AHC and CHB. Let's write down expressions for trigonometric functions:
From this it is clear that . Adding, we get the Pythagorean theorem, since:
For another proof of the Pythagorean theorem, see the commentary to Problem 4.
3 An important example of an additional construction is the construction of an angle equal to one of the angles of a triangle.
From the vertex of the right angle we draw a straight line segment that forms an angle with leg CA equal to angle CAB of the given right triangle ABC. As a result, we obtain an isosceles triangle ACM with base angles. But the other triangle resulting from this construction will also be isosceles, since each of its angles at the base is equal (by the property of the angles of a right triangle and by construction - the angle was “subtracted” from the right angle). Due to the fact that triangles BMC and AMC are isosceles with common side MC, we have the equality MB=MA=MC, i.e. M.C. median drawn to the hypotenuse of a right triangle and she equal to half the hypotenuse.
Corollary 1. The midpoint of the hypotenuse is the center of the circle circumscribed around this triangle, since it turns out that the midpoint of the hypotenuse is equidistant from the vertices of the right triangle.
Corollary 2. Middle line of a right triangle, connecting the middle of the hypotenuse and the middle of the leg, is parallel to the opposite leg and is equal to its half.
In isosceles triangles BMC and AMC, let us lower the heights MH and MG to the bases. Since in isosceles triangle, the height lowered to the base is also the median (and bisector), then MH and MG are the lines of a right triangle connecting the middle of the hypotenuse with the midpoints of the legs. By construction, they turn out to be parallel to the opposite legs and equal to their halves, since the triangles are equal MHC and MGC are equal (and MHCG is a rectangle). This result is the basis for the proof of the theorem on the midline of an arbitrary triangle and, further, the midline of a trapezoid and the property of proportionality of segments cut off by parallel lines on two straight lines intersecting them.
Tasks
Using similarity properties -1
Using basic properties - 2
Using additional formation 3-4
1 2 3 4
The height dropped from the vertex of a right angle of a right triangle is equal to the square root of the lengths of the segments into which it divides the hypotenuse.
The solution seems obvious if you know the derivation of the Pythagorean theorem from the similarity of triangles:
\(\mathrm(tg)\beta=\frac(h)(c_1)=\frac(c_2)(h)\),
whence \(h^2=c_1c_2\).
Find the locus of points (GMT) of intersection of the medians of all possible right triangles whose hypotenuse AB is fixed.
The point of intersection of the medians of any triangle cuts off one third from the median, counting from the point of its intersection with the corresponding side. IN right triangle The median drawn from a right angle is equal to half the hypotenuse. Therefore, the desired GMT is a circle of radius equal to 1/6 of the length of the hypotenuse, with a center in the middle of this (fixed) hypotenuse.
The midline of a trapezoid, and especially its properties, are very often used in geometry to solve problems and prove certain theorems.
is a quadrilateral with only 2 sides parallel to each other. The parallel sides are called bases (in Figure 1 - AD And B.C.), the other two are lateral (in the figure AB And CD).
Midline of trapezoid is a segment connecting the midpoints of its sides (in Figure 1 - KL).
Properties of the midline of a trapezoid
Proof of the trapezoid midline theorem
Prove that the midline of a trapezoid is equal to half the sum of its bases and is parallel to these bases.
Given a trapezoid ABCD with midline KL. To prove the properties under consideration, it is necessary to draw a straight line through the points B And L. In Figure 2 this is a straight line BQ. And also continue the foundation AD to the intersection with the line BQ.
Consider the resulting triangles L.B.C. And LQD:
- By definition of the midline KL dot L is the midpoint of the segment CD. It follows that the segments C.L. And LD are equal.
- ∠BLC = ∠QLD, since these angles are vertical.
- ∠BCL = ∠LDQ, since these angles lie crosswise on parallel lines AD And B.C. and secant CD.
From these 3 equalities it follows that the previously considered triangles L.B.C. And LQD equal on 1 side and two adjacent angles (see Fig. 3). Hence, ∠LBC = ∠ LQD, BC=DQ and most importantly - BL=LQ => KL, which is the midline of the trapezoid ABCD, is also the midline of the triangle ABQ. According to the property of the midline of a triangle ABQ we get.
\[(\Large(\text(Similarity of triangles)))\]
Definitions
Two triangles are called similar if their angles are respectively equal and the sides of one triangle are proportional to the similar sides of the other
(sides are called similar if they lie opposite equal angles).
The coefficient of similarity of (similar) triangles is a number equal to the ratio of the similar sides of these triangles.
Definition
The perimeter of a triangle is the sum of the lengths of all its sides.
Theorem
The ratio of the perimeters of two similar triangles is equal to the similarity coefficient.
Proof
Consider triangles \(ABC\) and \(A_1B_1C_1\) with sides \(a,b,c\) and \(a_1, b_1, c_1\) respectively (see figure above).
Then \(P_(ABC)=a+b+c=ka_1+kb_1+kc_1=k(a_1+b_1+c_1)=k\cdot P_(A_1B_1C_1)\)
Theorem
The ratio of the areas of two similar triangles is equal to the square of the similarity coefficient.
Proof
Let the triangles \(ABC\) and \(A_1B_1C_1\) be similar, and \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1) = k\). Let us denote by the letters \(S\) and \(S_1\) the areas of these triangles, respectively.
Since \(\angle A = \angle A_1\) , then \(\dfrac(S)(S_1) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\)(by the theorem on the ratio of the areas of triangles having equal angles).
Because \(\dfrac(AB)(A_1B_1) = \dfrac(AC)(A_1C_1) = k\), That \(\dfrac(S)(S_1) = \dfrac(AB)(A_1B_1)\cdot\dfrac(AC)(A_1C_1) = k\cdot k = k^2\), which was what needed to be proven.
\[(\Large(\text(Signs of similarity of triangles)))\]
Theorem (the first sign of similarity of triangles)
If two angles of one triangle are respectively equal to two angles of another triangle, then such triangles are similar.
Proof
Let \(ABC\) and \(A_1B_1C_1\) be triangles such that \(\angle A = \angle A_1\) , \(\angle B = \angle B_1\) . Then, by the theorem on the sum of angles of a triangle \(\angle C = 180^\circ - \angle A - \angle B = 180^\circ - \angle A_1 - \angle B_1 = \angle C_1\), that is, the angles of the triangle \(ABC\) are respectively equal to the angles of the triangle \(A_1B_1C_1\) .
Since \(\angle A = \angle A_1\) and \(\angle B = \angle B_1\) , then \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot AC)(A_1B_1\cdot A_1C_1)\) And \(\dfrac(S_(ABC))(S_(A_1B_1C_1)) = \dfrac(AB\cdot BC)(A_1B_1\cdot B_1C_1)\).
From these equalities it follows that \(\dfrac(AC)(A_1C_1) = \dfrac(BC)(B_1C_1)\).
Similarly, it is proved that \(\dfrac(AC)(A_1C_1) = \dfrac(AB)(A_1B_1)\)(using equalities \(\angle B = \angle B_1\) , \(\angle C = \angle C_1\) ).
As a result, the sides of the triangle \(ABC\) are proportional to the similar sides of the triangle \(A_1B_1C_1\), which is what needed to be proven.
Theorem (second criterion for the similarity of triangles)
If two sides of one triangle are proportional to two sides of another triangle and the angles between these sides are equal, then the triangles are similar.
Proof
Consider two triangles \(ABC\) and \(A"B"C"\) such that \(\dfrac(AB)(A"B")=\dfrac(AC)(A"C")\), \(\angle BAC = \angle A"\) Let us prove that the triangles \(ABC\) and \(A"B"C"\) are similar. Taking into account the first sign of similarity of triangles, it is enough to show that \(\angle B = \angle B"\) .
Consider a triangle \(ABC""\) with \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) . Triangles \(ABC""\) and \(A"B"C"\) are similar according to the first criterion of similarity of triangles, then \(\dfrac(AB)(A"B") = \dfrac(AC"")(A"C")\).
On the other hand, by condition \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C")\). From the last two equalities it follows that \(AC = AC""\) .
Triangles \(ABC\) and \(ABC""\) are equal in two sides and the angle between them, therefore, \(\angle B = \angle 2 = \angle B"\).
Theorem (third sign of similarity of triangles)
If three sides of one triangle are proportional to three sides of another triangle, then the triangles are similar.
Proof
Let the sides of the triangles \(ABC\) and \(A"B"C"\) be proportional: \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\). Let us prove that the triangles \(ABC\) and \(A"B"C"\) are similar.
To do this, taking into account the second criterion for the similarity of triangles, it is enough to prove that \(\angle BAC = \angle A"\) .
Consider a triangle \(ABC""\) with \(\angle 1 = \angle A"\) , \(\angle 2 = \angle B"\) .
Triangles \(ABC""\) and \(A"B"C"\) are similar according to the first criterion of similarity of triangles, therefore, \(\dfrac(AB)(A"B") = \dfrac(BC"")(B"C") = \dfrac(C""A)(C"A")\).
From the last chain of equalities and conditions \(\dfrac(AB)(A"B") = \dfrac(AC)(A"C") = \dfrac(BC)(B"C")\) it follows that \(BC = BC""\) , \(CA = C""A\) .
Triangles \(ABC\) and \(ABC""\) are equal on three sides, therefore, \(\angle BAC = \angle 1 = \angle A"\).
\[(\Large(\text(Thales' Theorem)))\]
Theorem
If you mark equal segments on one side of an angle and draw parallel straight lines through their ends, then these straight lines will also cut off equal segments on the other side.
Proof
Let's prove first lemma: If in \(\triangle OBB_1\) a straight line \(a\parallel BB_1\) is drawn through the middle \(A\) of side \(OB\), then it will also intersect side \(OB_1\) in the middle.
Through the point \(B_1\) we draw \(l\parallel OB\) . Let \(l\cap a=K\) . Then \(ABB_1K\) is a parallelogram, therefore \(B_1K=AB=OA\) and \(\angle A_1KB_1=\angle ABB_1=\angle OAA_1\); \(\angle AA_1O=\angle KA_1B_1\) like vertical. So, according to the second sign \(\triangle OAA_1=\triangle B_1KA_1 \Rightarrow OA_1=A_1B_1\). The lemma is proven.
Let's move on to the proof of the theorem. Let \(OA=AB=BC\) , \(a\parallel b\parallel c\) and we need to prove that \(OA_1=A_1B_1=B_1C_1\) .
Thus, according to this lemma \(OA_1=A_1B_1\) . Let's prove that \(A_1B_1=B_1C_1\) . Let us draw a line \(d\parallel OC\) through the point \(B_1\), and let \(d\cap a=D_1, d\cap c=D_2\) . Then \(ABB_1D_1, BCD_2B_1\) are parallelograms, therefore, \(D_1B_1=AB=BC=B_1D_2\) . Thus, \(\angle A_1B_1D_1=\angle C_1B_1D_2\) like vertical \(\angle A_1D_1B_1=\angle C_1D_2B_1\) lying like crosses, and, therefore, according to the second sign \(\triangle A_1B_1D_1=\triangle C_1B_1D_2 \Rightarrow A_1B_1=B_1C_1\).
Thales's theorem
Parallel lines cut off proportional segments on the sides of an angle.
Proof
Let parallel lines \(p\parallel q\parallel r\parallel s\) divided one of the lines into segments \(a, b, c, d\) . Then the second straight line should be divided into segments \(ka, kb, kc, kd\), respectively, where \(k\) is a certain number, the same proportionality coefficient of the segments.
Let us draw through the point \(A_1\) a line \(p\parallel OD\) (\(ABB_2A_1\) is a parallelogram, therefore, \(AB=A_1B_2\) ). Then \(\triangle OAA_1 \sim \triangle A_1B_1B_2\) at two corners. Hence, \(\dfrac(OA)(A_1B_2)=\dfrac(OA_1)(A_1B_1) \Rightarrow A_1B_1=kb\).
Similarly, we draw a straight line through \(B_1\) \(q\parallel OD \Rightarrow \triangle OBB_1\sim \triangle B_1C_1C_2 \Rightarrow B_1C_1=kc\) etc.
\[(\Large(\text(Middle line of the triangle)))\]
Definition
The midline of a triangle is a segment connecting the midpoints of any two sides of the triangle.
Theorem
The middle line of the triangle is parallel to the third side and equal to half of it.
Proof
1) The parallelism of the midline to the base follows from what was proven above lemmas.
2) Let us prove that \(MN=\dfrac12 AC\) .
Through the point \(N\) we draw a line parallel to \(AB\) . Let this line intersect the side \(AC\) at the point \(K\) . Then \(AMNK\) is a parallelogram ( \(AM\parallel NK, MN\parallel AK\) according to the previous point). So, \(MN=AK\) .
Because \(NK\parallel AB\) and \(N\) are the midpoint of \(BC\), then by Thales’ theorem \(K\) is the midpoint of \(AC\) . Therefore, \(MN=AK=KC=\dfrac12 AC\) .
Consequence
The middle line of the triangle cuts off from it a triangle similar to the given one with the coefficient \(\frac12\) .
Middle line figures in planimetry - a segment connecting the midpoints of two sides of a given figure. The concept is used for the following figures: triangle, quadrilateral, trapezoid.
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Subtitles
Middle line of the triangle
Properties
- the middle line of the triangle is parallel to the base and equal to half of it.
- when all three middle lines intersect, 4 are formed equal triangle, similar (even homothetic) to the original one with a coefficient of 1/2.
- the middle line cuts off a triangle that is similar to this one, and its area is equal to one-fourth the area of the original triangle.
- The three middle lines of the triangle divide it into 4 equal (identical) triangles, similar to the original triangle. All 4 such identical triangles are called medial triangles. The central one of these 4 identical triangles is called the complementary triangle.
Signs
- if a segment is parallel to one of the sides of the triangle and connects the midpoint of one side of the triangle to a point lying on the other side of the triangle, then it is a midline.
Midline of a quadrilateral
Midline of a quadrilateral- a segment connecting the midpoints of opposite sides of a quadrilateral.
Properties
The first line connects 2 opposite sides. The second connects the other 2 opposite sides. The third connects the centers of two diagonals (not in all quadrilaterals the diagonals are divided in half at the point of intersection).
- If in a convex quadrilateral the middle line forms equal angles with the diagonals of the quadrilateral, then the diagonals are equal.
- The length of the midline of a quadrilateral is less than half the sum of the other two sides or equal to it if these sides are parallel, and only in this case.
- The midpoints of the sides of an arbitrary quadrilateral are the vertices of a parallelogram. Its area is equal to half the area of the quadrilateral, and its center lies at the point of intersection of the middle lines. This parallelogram is called Varignon's parallelogram;
- The last point means the following: In a convex quadrilateral you can draw four midlines of the second kind. Midlines of the second kind- four segments inside a quadrilateral passing through its midpoints adjacent sides parallel to the diagonals. Four midlines of the second kind of a convex quadrilateral, cut it into four triangles and one central quadrilateral. This central quadrilateral is a Varignon parallelogram.
- The point of intersection of the midlines of a quadrilateral is their common midpoint and bisects the segment connecting the midpoints of the diagonals. Moreover, she is
You are wondering how you can calculate and find the midline of a triangle. Then let's get to work.
Finding the length of the midline of a triangle is quite simple. Since a triangle has three sides, there are accordingly three angles, and it is possible that when constructing three middle lines.
What is a triangle?
Three sides (equilateral, isosceles)
Three angles (acute, obtuse, right triangles, respectively)
What is the midline of a triangle
This is a segment. A line segment connects the midpoint of two sides of a triangle. Any triangle has three middle lines.
Property 1: The midline of a triangle is parallel to a side of the triangle and equal to its half. Therefore, to determine the midline of a triangle, it is enough to know the length of the third side.
Example: yes triangle ABC, it is known that the middle side KN is parallel to AC. Length AC = 8 cm, AB = 4 cm, BC = 4 cm. Therefore, to find the midline of the triangle, AC/2 is sufficient to obtain the midline of the triangle. Answer: 4 cm midline in a given triangle according to existing parameters.
Property 2: If three middle lines are drawn in a triangle, then four equal similar triangles are formed. The coefficient is ½.
Property 3: Midline equilateral triangle splits the triangle into a trapezoid and a triangle.
An example of solving a problem: If we draw a triangle, we will see that at the top of the triangle there is a figure with three corners. At the bottom of the quadrilateral is a figure with two opposite sides, which are parallel to each other.
