Proportional segments in a circle. Maths. Algebra. Geometry. Trigonometry

Theorem 111. 1) The perpendicular, dropped from some point on the circle to the diameter, is mid-proportional between the parts of the diameter. This perpendicular is sometimes called the ordinate.

2) The chord connecting the end of the diameter with the point of the circle is mid-proportional between the diameter and the segment adjacent to the chord.

Given. Let us drop from some point C of the circle the perpendicular CD to the diameter AB (Fig. 169).

It is required to prove that 1) AD/CD = CD/DB, and 2) AD/AC = AC/AB.

Proof. Let us join the point C with the ends of the diameter AB, then at the point C a right angle ACB is formed, in which the segment CD is a perpendicular dropped from the vertex of the right angle to the hypotenuse.

Based on Theorem 100, the following proportion holds:

based on theorem 101, the proportion is:

AD/AC = AC/AB, DB/CB = CB/AB (1)

Consequence. The squares of the chords are related as the corresponding segments of the diameter.

Proof. Equations follow from proportion (1):

AC 2 = AB AD, CB 2 = AB BD

whence by separation we find:

AC 2 /CB 2 = AD/DB.

Theorem 112. The parts of intersecting chords are inversely proportional to each other.

Two intersecting chords AB and CD are given (Fig. 170).

It is required to prove that

i.e. the greater part of the first chord is related to the greater part of the second as the lesser part of the second chord is to the lesser part of the first.

Proof. Let us connect point A with C and B with D, then two similar triangles ACE and DBE are formed, because the angles at point E are equal as vertical, ∠CAB = ∠CDB as leaning on the ends of the arc CB, ∠ACD = ∠ABD as leaning on the ends of the arc AD.

From the similarity of triangles ACE and DBE follows the proportion:

BE/DE = CE/AE (a)

Proportion (a) implies the equality:

BE AE = DE CE

showing that the product of the segments of one is equal to the product of the segments of the other chord.

Theorem 113. Two secants drawn from the same point outside the circle are inversely proportional to their outer parts.

Given two secants AB and AC drawn from point A (Drawing 171).

It is required to prove that

i.e., the first secant is related to the second, as the outer part of the second is related to the outer part of the first secant.

Proof. Connect points D to C and B to E.

The two triangles ∠ABE and ∠ADC are similar, because the angle A is common, B = C as based on the ends of the same arc DE, hence ∠ADC = ∠AEB.

From the similarity of triangles ADC and ABE follows the proportion:

AC / AB \u003d AD / AE (PTD).

From the same proportion follows the equality

AC AE = AB AD

showing that the product of a secant and its outer segment is equal to the product of another secant and its segment(if the secants come from the same point).

Theorem 114. The tangent is mid-proportional between the whole secant and its outer part.

Tangent AB and secant BC are given (Fig. 172).

It is required to prove that

Proof. Let's connect point A with points C and D.

Triangles ABC and ABD are similar because angle B is common, ∠BAD = ∠ACD, therefore ∠CAB = ∠ADB.

BC / AB \u003d AB / BD (PTD).

From this proportion follows the equality:

AB 2 = BC BD

showing that the square of the tangent is equal to the product of the secant and its outer part.

Property of the sides of an inscribed quadrilateral

Theorem 115. In any quadrilateral inscribed in a circle, the product of the diagonals is equal to the sum of the products of the opposite sides.

This assumption, known as Ptolemy's theorem, occurs for the first time in Ptolemy's Alagegest in the 2nd century A.D.

An inscribed quadrilateral ABCD is given (Fig. 173) and diagonals AC and BD are drawn.

It is required to prove that AC · BD = AB · CD + BC · AD.

Proof. Draw a line BE so that the angle EBC is equal to the angle ABD. Two triangles ABD and BEC are similar, because ∠ABD = ∠CBE by construction, ∠ADB = ∠BCE as based on the same arc AB, therefore,

From the similarity of these triangles follows the proportion:

BC/BD = EC/AD (a)

Triangles ABE and BCD are similar, because ∠ABE = ∠DBC by construction, ∠BAE = ∠BDC as based on the arc BC, therefore,

∠BEA = ∠BCD.

From the similarity of these triangles follows the proportion:

AB/BD = AE/CD (b)

Proportions (a) and (b) imply equalities:

BC AD = BD EC
AB CD = BD AE

Adding these equalities, we have:

BC AD + AB CD = BD EC + BD AE = BD (EC + AE)

Since EC + AE = AC, then

BD AC = BC AD + AB CD (PTD).

Theorem 116. In any inscribed quadrilateral, the diagonals are related as the sums of the products of the sides based on the ends of the diagonals.

An inscribed quadrilateral ABCD is given (Fig. 174) and diagonals AC and BD are drawn.

It is required to prove that

BD/AC = (AD DC + AB BC) / (BC CD + AD AB)

Proof. a) From point B we set aside the arc BE equal to DC and connect point E with points A, B, D.

For the inscribed quadrilateral ABED, the equality holds:

AE BD = AD BE + AB DE.

Since BE = CD by construction, DE = BC because ◡DE = ◡DC + ◡CE and ◡BC = ◡BE + ◡CE.

Replacing BE and DE with their values, we have the equality:

AE BD = AD CD + AB BC (a)

b) Setting aside the arc AF from the point A equal to the arc BC and connecting the point F with the points A, D, C, we have the equality for the quadrilateral AFCD:

AC DF = AF CD + AD CF

In this equality, AF = BC by construction, CF = AB (because ◡CF = ◡BC + ◡BF and ◡AB = ◡AF + ◡BF = ◡BC + ◡BF)

Replacing the quantities AF and CF with their quantities, we find the equality:

AC DF = BC CD + AD AB (b)

In equalities (a) and (b), the segments AE and DF are equal, because

◡ADE = AD + DE = ◡AD + ◡BC = ◡AD + ◡AF = ◡DAF

Separating equalities (a) and (b), we find:

BC/AD = (AD C D + AB BC) / (BC CD + AD AB)(ChTD).

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 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.

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 indicated 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.