X. proportional segments in a right triangle and a circle. trigonometric functions of an acute angle. Additional properties

Let us first consider the secant AC, drawn from the point A external to the given circle (Fig. 288). Draw the tangent AT from the same point. We will call the segment between point A and the point of intersection closest to it with the circle the outer part of the secant (segment AB in Fig. 288), while the segment AC to the farthest of the two intersection points is simply the secant. The tangent segment from A to the point of contact is also briefly called the tangent. Then

Theorem. The product of a secant and its outer part is equal to the square of the tangent.

Proof. Let's connect the dot. Triangles ACT and BT A are similar, since they have a common angle at vertex A, and angles ACT and are equal, since both of them are measured by half of the same arc TB. Therefore, from here we get the required result:

The tangent is equal to the geometric mean between the secant drawn from the same point and its outer part.

Consequence. For any secant drawn through a given point A, the product of its length and the outer part is constant:

Consider now chords intersecting at an interior point. Correct statement:

If two chords intersect, then the product of the segments of one chord is equal to the product of the segments of the other (meaning the segments into which the chord is divided by the intersection point).

So, in fig. 289 the chords AB and CD intersect at the point M, and we have In other words,

For a given point M, the product of the segments into which it divides any chord passing through it is constant.

To prove this, we note that the triangles MBC and MAD are similar: the angles CMB and DMA are vertical, the angles MAD and MCB are based on the same arc. From here we find

Q.E.D.

If a given point M lies at a distance l from the center, then, drawing a diameter through it and considering it as one of the chords, we find that the product of the segments of the diameter, and hence of any other chord, is equal to It is also equal to the square of the minimum half-chord (perpendicular to specified diameter) passing through M.

The theorem on the constancy of the product of segments of a chord and the theorem on the constancy of the product of a secant by its outer part are two cases of the same statement, the only difference is whether the secants are drawn through an external or internal point of the circle. Now you can specify one more feature that distinguishes inscribed quadrangles:

In any inscribed quadrilateral, the cutoff products into which the diagonals are divided by their intersection point are equal.

The necessity of the condition is obvious, since the diagonals will be the chords of the circumscribed circle. It can be shown that this condition is also sufficient.

Maths. Algebra. Geometry. Trigonometry

GEOMETRY: Planimetry

10. Theorems on proportional lines

Theorem. The sides of the angle are intersected by a number of parallel lines, cut by them into proportional parts.

Proof. It is required to prove that

.

Drawing auxiliary lines DM,EN,... parallel to BA, we get triangles that are similar to each other, since their angles are respectively equal (due to the parallelism of the lines). From their similarity it follows:

Replacing the segment DM with D"E" in this series of equal ratios, the segment EN with E"F" (opposite sides of the parallelogram), we get what we wanted to prove.

Theorem. The bisector of any angle of a triangle divides the opposite side into parts proportional to the adjacent sides of the triangle

.

Inverse theorem. If any side of a triangle is divided into two parts proportional to two adjacent sides of this triangle, then the line connecting the point of division with the vertex of the opposite angle is the bisector of this angle

.

Theorem. If the bisector of an external angle of a triangle intersects the extension of the opposite side at some point, then the distances from this point to the ends of the extended side are proportional to the adjacent sides of the triangle

.

Numerical dependencies between the elements of a triangle.

Theorem. In a right triangle, the perpendicular dropped from the vertex of the right angle to the hypotenuse is the average proportional between the segments of the hypotenuse, and each leg is the average proportional between the hypotenuse and the segment adjacent to this leg

.

Proof. It is required to prove the following three proportions: 1) BD:AD=AD:DC, 2) BC:AB=AB:DB, 3) BC:AC=AC:DC.

1) Triangles ABD and ADC are similar because

P 1=P 4 and P 2=P 3 (since their sides are perpendicular), hence BD:AD=AD:DC.

2) Triangles ABD and ABC are similar, since they are right-angled and have a common angle B, hence BC:AB=AB:DB.

3) Triangles ABC and ADC are similar, since they are rectangular and have a common angle C, therefore BC:AC=AC:DC.

Consequence. The perpendicular dropped from some point on the circle to the diameter is the average proportional between the segments of the diameter, and the chord connecting this point with the end of the diameter is the average proportional between the diameter and the segment adjacent to the chord

.

Pythagorean theorem. In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the legs

.

Consequence. The squares of the legs are related to each other as adjacent segments of the hypotenuse

.

Theorem. In any triangle, the square of the side opposite the acute angle is equal to the sum of the squares of the other two sides without double

product of any of these sides by its segment from the apex of an acute angle to a height.

Theorem. The sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides

.

Proportional lines in a circle.

Theorem. If a chord and a diameter are drawn through a point taken inside the circle, then the product of the segments of the chord is equal to the product of the segments of the diameter.

Consequence. If any number of chords are drawn through a point taken inside the circle, then the product of the segments of each chord is a constant number for all chords.

Theorem. If some secant and tangent are drawn to it from a point taken outside the circle, then the product of the secant and its outer part is equal to the square of the tangent

.

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§ 11. Proportional segments in a circle.

1. The bridge truss is bounded by an arc of a circle (Fig. 38); truss height MK= h= 3 m; arc radius AMB of the span R = 8.5 m. Calculate the length AB of the bridge span.

2. In a semi-cylinder vaulted basement, two posts should be placed, each at the same distance from the nearest wall. Determine the height of the racks if the width of the basement along the bottom is 4 m, and the distance between the racks is 2 m.

3. 1) A perpendicular to the diameter is drawn from the point of the circle. Determine its length with the following length of segments of diameter: 1) 12 cm and 3 cm; 2) 16 cm and 9 cm, 3) 2 m and 5 dm.

2) A perpendicular is drawn from the point of the diameter to the intersection with the circle. Determine the length of this perpendicular if the diameter is 40 cm, and the drawn perpendicular is 8 cm from one of the ends of the diameter.

4. The diameter is divided into segments: AC \u003d 8 dm and CB \u003d 5 m, and from point C a perpendicular CD of a given length is drawn to it. Indicate the position of point D relative to the circle when CD equals: 1) 15 dm; 2) 2 m; 3) 23 dm.

5. DIA-semicircle; CD - perpendicular to the diameter AB. Required:

1) determine DB if AD = 25 and CD =10;

2) determine AB if AD: DB= 4: 9 and CD=30;

3) define AD if CD=3AD and the radius is r;

4) determine AD if AB=50 and CD=15.

6. 1) The perpendicular, lowered from the point of the circle to a radius of 34 cm, divides it in a ratio of 8:9 (starting from the center). Determine the length of the perpendicular.

2) The chord BDC is perpendicular to the radius ODA. Determine BC if OA = 25 cm and AD = 10 cm.

3) The width of the ring formed by two concentric circles is 8 dm; the chord of the larger circle, tangent to the smaller one, is 4 m. Determine the radii of the circles.

7. By comparing the segments, prove that the arithmetic mean of two unequal numbers is greater than their geometric mean.

8. Construct a segment, the average proportional between the segments 3 cm and 5 cm.

9. Construct a segment equal to: √15; √10; √6; √3.

10. ADB-diameter; AC-chord; CD is perpendicular to the diameter. Determine the AC chord: 1) if AB = 2 m and AD = 0.5 m; 2) if AD = 4 cm and DB = 5 cm; 3) if AB=20m and DB=15m.

11. AB diameter; AC-chord; AD is its projection on the diameter AB. Required:

1) determine AD if AB=18 cm and AC=12 cm;

2) determine the radius if AC=12 m and AD=4 m;

3) determine DB if AC=24 cm and DB = 7/9 AD.

12. AB diameter; AC-chord; AD is its projection on the diameter AB. Required:

1) determine AC if AB = 35 cm and AC=5AD;

2) determine AC if the radius is equal to r and AC=DB.

13. Two chords intersect inside a circle. The segments of one chord are 24 cm and 14 cm; one of the segments of the other chord is 28 cm. Determine its second segment.

14. The bridge truss is limited by an arc of a circle (Fig. 38); bridge length AB = 6 m, height A = 1.2 m. Determine the radius of the arc (OM = R).

15. Two segments AB and CD intersect at point M so that MA \u003d 7 cm, MB \u003d 21 cm,
MC = 3 cm and MD = 16 cm. Do points A, B, C and D lie on the same circle?

16. Pendulum length MA = l= 1 m (Fig. 39), its lifting height, when deviated by an angle α, CA = h\u003d 10 cm. Find the distance BC of point B from MA (BC \u003d X).

17. To translate the railway track width b\u003d 1.524 m in place AB (Fig. 40) a rounding is made; while it turned out, ; that BC= a= 42.4 m. Determine the radius of curvature OA = R.

18. The chord AMB is rotated near the point M so that the segment MA has increased 2 1/2 times. How has the segment MB changed?

19. 1) Of the two intersecting chords, one was divided into parts of 48 cm and 3 cm, and the other in half. Determine the length of the second chord.

2) Of the two intersecting chords, one was divided into parts of 12 m and 18 m, and the other in a ratio of 3:8. Determine the length of the second chord.

20. Of the two intersecting chords, the first is 32 cm, and the segments of the other chord are
12 cm and 16 cm. Determine the segments of the first chord.

21. The secant ABC is rotated near the outer point A so that its outer segment AB has decreased three times. How did the length of the secant change?

22. Let ADB and AEC be two lines intersecting the circle: the first is at points D and B, the second is at points E and C. Required:

1) determine AE if AD = 5 cm, DB = 15 cm and AC = 25 cm;

2) determine BD if AB = 24 m, AC = 16 m and EC = 10 m;

3) determine AB and AC if AB+AC=50 m, a AD: AE = 3:7.

23. The radius of the circle is 7 cm. From a point 9 cm away from the center, a secant is drawn so that it is divided in half by the circle. Determine the length of this secant.

24. MAB and MCD are two secants to one circle. Required:

1) determine CD if MV = 1 m, MD = 15 dm and CD = MA;

2) determine MD if MA =18 cm, AB=12 cm and MC:CD = 5:7;

3) determine AB if AB=MC, MA=20 and CD=11.

25. Two chords are extended to mutual intersection. Determine the length of the resulting extensions if the chords are equal a and b, and their extensions are related as t:p.

26. From one point, a secant and a tangent are drawn to the circle. Determine the length of the tangent if the outer and inner segments of the secant are respectively expressed by the following numbers: 1) 4 and 5; 2) 2.25 and 1.75; 3) 1 and 2.

27. The tangent is 20 cm, and the largest secant drawn from the same point is 50 cm. Determine the radius of the circle.

28. The secant is 2 1/4 times larger than its outer segment. How many times greater is it than a tangent drawn from the same point?

29. The common chord of two intersecting circles is continued, and tangents are drawn to them from a point taken on the continuation. Prove that they are equal.

30. On one side of the corner A, segments are laid one after the other: AB \u003d 6 cm and BC \u003d 8 cm; and on the other side a segment AD = 10 cm is laid down. A circle is drawn through points B, C and D. Find out if the line AD touches this circle, and if not, then whether the point D will be the first (counting from A) or the second point of intersection.

31. Let it be: AB-tangent and ACD-secant of the same circle. Required:

1) determine CD if AB = 2 cm and AD = 4 cm;

2) determine AD if AC:CD = 4:5 and AB=12 cm;

3) determine AB if AB = CD and AC = a.

32. 1) How far can you see from a balloon (Fig. 41), which has risen to a height of 4 km above the ground (the radius of the earth is = 6370 km)?

2) Mount Elbrus (in the Caucasus) rises 5,600 meters above sea level. How far can you see from the top of this mountain?

3) M - an observation post with a height of A meters above the ground (Fig. 42); earth radius R, МТ= d is the greatest visible distance. Prove that d= √2R h+ h 2

Comment. Because h 2 due to its smallness compared to 2R h almost does not affect the result, then you can use the approximate formula d≈ √2R h .

33. 1) Tangent and secant, coming out of one point, are respectively equal to 20 cm and 40 cm; the secant is 8 cm away from the center. Determine the radius of the circle.

2) Determine the distance from the center to the point from which the tangent and secant go, if they are respectively 4 cm and 8 cm, and the secant is removed from the center by
12 cm

34. 1) From a common point, a tangent and a secant are drawn to the circle. Determine the length of the tangent if it is 5 cm longer than the outer segment of the secant and by the same amount less than the inner segment.

2) From one point, a secant and a tangent are drawn to the circle. The secant is a, and its inner segment is longer than the outer segment by the length of the tangent. Define tangent.

36. From one point, a tangent and a secant are drawn to one circle. The tangent is greater than the inner and outer segments of the secant by 2 cm and 4 cm, respectively. Determine the length of the secant.

36. From one point, a tangent and a secant are drawn to the circle. Determine their length if the tangent is 20 cm less than the inner segment of the secant and 8 cm more than the outer segment.

37. 1) A secant and a tangent are drawn from one point to the circle. Their sum is 30 cm, and the inner segment of the secant is 2 cm less than the tangent. Define secant and tangent.

2) From one point, a secant and a tangent are drawn to the circle. Their sum is 15 cm, and the outer segment of the secant is 2 cm less than the tangent. Define secant and tangent.

38. The segment AB is extended by the distance BC. On AB and AC, as on diameters, circles are built. A perpendicular BD is drawn to the segment AC at point B until it intersects with a larger circle. From point C, a tangent SC is drawn to the smaller circle. Prove that CD = CK.

39. Two parallel tangents and a third tangent that intersect them are drawn to a given circle. The radius is the average proportional between the segments of the third tangent. Prove.

40. Two parallel lines are given at a distance of 15 dm from one another; a point M is given between them at a distance of 3 dm from one of them. A circle is drawn through the point M, tangent to both parallels. Determine the distance between the projections of the center and point M on one of these parallels.

41. In a circle of radius r An isosceles triangle is inscribed in which the sum of the height and base is equal to the diameter of the circle. Determine height.

42. Determine the radius of a circle circumscribed about an isosceles triangle: 1) if the base is 16 cm and the height is 4 cm; 2) if the side is 12 dm and the height is 9 dm; 3) if the side is 15 m and the base is 18 m.

43. In an isosceles triangle, the base is 48 dm, and the side is 30 dm. Determine the radii of the circles, circumscribed and inscribed, and the distance between their centers.

44. The radius is r, the chord of this arc is equal to a. Determine the chord of the doubled arc.

45. The radius of the circle is 8 dm; chord AB is 12 dm. A tangent is drawn through point A, and from point B is a chord BC parallel to the tangent. Determine the distance between the tangent and the chord BC.

46. ​​Point A is removed from the straight line MN at a distance With. given radius r A circle is circumscribed so that it passes through point A and touches line MN. Determine the distance between the received point of contact and the given point A.

Property 1 . If the chords AB and CD of the circle intersect at point S, then AS BS = CS DS, i.e. DS/BS = AS/CS.

Proof. Let us first prove that the triangles ASD and CSB are similar.

The inscribed angles DCB and DAB are equal as they are based on the same arc.

Angles ASD and BSC are equal as vertical.

From the equality of the indicated angles it follows that the triangles ASD and CSB are similar. From the similarity of triangles follows the proportion

DS/BS = AS/CS, or AS BS = CS DS,

Q.E.D.

Property 2. If two secants are drawn from the point P to the circle, intersecting the circle at points A, B and C, D, respectively, then АР/СР = DP/BP.

Proof. Let A and C be the points of intersection of the secants with the circle closest to the point P. Triangles PAD and RSV are similar. They have a common angle at the vertex P, and the angles B and D are equal as inscribed, based on the same arc. From the similarity of triangles follows the proportion АР/СР = DP/BP, which was required to be proved.

Bisector property of an angle of a triangle

The angle bisector of a triangle divides the opposite side into segments proportional to the other two sides.

Proof. Let CD be the bisector of triangle ABC. If the triangle ABC is isosceles with base AB, then the specified property of the bisector is obvious, since in this case the bisector is also the median. Consider the general case where AC is not equal to BC. Let us drop perpendiculars AF and BE from vertices A and B to line CD. Right triangles ACF and ALL are similar, since they have equal acute angles at vertex C.

From the similarity of triangles, the proportionality of the sides follows: AC / BC \u003d AF / BE. Right triangles ADF and BDE are also similar. Their angles at the vertex D are equal as vertical. It follows from the similarity: AF/BE = AD/BD. Comparing this equality with the previous one, we get: AC / BC \u003d AD / BD or AC / AD \u003d BC / BD, that is, AD and BD are proportional to the sides AC and BC.

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