INSCRIBED AND CIRCULAR POLYGONS,
§ 106. PROPERTIES OF INSCRIBED AND DESCRIBED QUADRIAGONS.
Theorem 1. Sum opposite corners cyclic quadrilateral is equal to 180°.
Let a quadrilateral ABCD be inscribed in a circle with center O (Fig. 412). It is required to prove that / A+ / C = 180° and / B + / D = 180°.
/
A, as inscribed in circle O, measures 1/2 BCD.
/
C, as inscribed in the same circle, measures 1/2 BAD.
Consequently, the sum of angles A and C is measured by the half-sum of arcs BCD and BAD; in sum, these arcs make up a circle, i.e. they have 360°.
From here /
A+ /
C = 360°: 2 = 180°.
Similarly, it is proved that / B + / D = 180°. However, this can be deduced in another way. We know that the amount internal corners convex quadrilateral equal to 360°. The sum of angles A and C is equal to 180°, which means that the sum of the other two angles of the quadrilateral also remains 180°.
Theorem 2(reverse). If in a quadrilateral the sum of two opposite angles is equal 180° , then a circle can be described around such a quadrilateral.
Let the sum of the opposite angles of the quadrilateral ABCD be equal to 180°, namely
/
A+ /
C = 180° and /
B + /
D = 180° (drawing 412).
Let us prove that a circle can be described around such a quadrilateral.
Proof. Through any 3 vertices of this quadrilateral you can draw a circle, for example through points A, B and C. Where will point D be located?
Point D can occupy only one of next three positions: to be inside the circle, to be outside the circle, to be on the circumference of the circle.
Let's assume that the vertex is inside the circle and takes position D" (Fig. 413). Then in the quadrilateral ABCD" we will have:
/ B + / D" = 2 d.
Continuing side AD" to the intersection with the circle at point E and connecting points E and C, we obtain the cyclic quadrilateral ABCE, in which, by the direct theorem
/ B+ / E = 2 d.
From these two equalities it follows:
/
D" = 2 d - /
B;
/
E=2 d - /
B;
/ D" = / E,
but this cannot be, because / D", being external relative to triangle CD"E, must be greater than angle E. Therefore, point D cannot be inside the circle.
It is also proved that vertex D cannot take position D" outside the circle (Fig. 414).
It remains to recognize that vertex D must lie on the circumference of the circle, i.e., coincide with point E, which means that a circle can be described around the quadrilateral ABCD.
Consequences. 1. A circle can be described around any rectangle.
2. Around isosceles trapezoid can describe a circle.
In both cases, the sum of opposite angles is 180°.
Theorem 3. In the described quadrilateral the sums opposite sides are equal. Let the quadrilateral ABCD be described about a circle (Fig. 415), i.e., its sides AB, BC, CD and DA are tangent to this circle.
It is required to prove that AB + CD = AD + BC. Let us denote the points of tangency by the letters M, N, K, P. Based on the properties of tangents drawn to a circle from one point (§ 75), we have:
AR = AK;
VR = VM;
DN = DK;
CN = CM.
Let us add these equalities term by term. We get:
AR + BP + DN + CN = AK + VM + DK + SM,
i.e. AB + CD = AD + BC, which is what needed to be proven.
Exercises.
1. In an inscribed quadrilateral, two opposite angles are in the ratio 3:5,
and the other two are in the ratio 4:5. Determine the magnitude of these angles.
2. In the described quadrilateral, the sum of two opposite sides is 45 cm. The remaining two sides are in the ratio 0.2: 0.3. Find the length of these sides.
A quadrilateral is inscribed in a circle if all its vertices lie on the circle. Such a circle is circumscribed about a quadrilateral.
Just as not every quadrilateral can be described around a circle, not every quadrilateral can be inscribed in a circle.
A convex quadrilateral inscribed in a circle has the property that its opposite angles add up to 180°. So, if given a quadrilateral ABCD, in which angle A is opposite to angle C, and angle B is opposite to angle D, then ∠A + ∠C = 180° and ∠B + ∠D = 180°.
In general, if one pair of opposite angles of a quadrilateral adds up to 180°, then the other pair will add up to the same amount. This follows from the fact that in a convex quadrilateral the sum of the angles is always equal to 360°. In turn, this fact follows from the fact that convex polygons the sum of angles is determined by the formula 180° * (n – 2), where n is the number of angles (or sides).
You can prove the property of an inscribed quadrilateral in the following way. Let a quadrilateral ABCD be inscribed in circle O. We need to prove that ∠B + ∠D = 180°.
Angle B is inscribed in a circle. As is known, such an angle equal to half arc on which it rests. IN in this case angle B is supported by arc ADC, which means ∠B = ½◡ADC. (Since the arc is equal to the angle between the radii forming it, we can write that ∠B = ½∠AOC, the inner region of which contains point D.)
On the other side, angle D of the quadrilateral rests on arc ABC, that is, ∠D = ½◡ABC.
Since the sides of angles B and D intersect the circle at the same points (A and C), they divide the circle into only two arcs - ◡ADC and ◡ABC. Because full circle adds up to 360°, then ◡ADC + ◡ABC = 360°.
Thus, the following equalities were obtained:
∠B = ½◡ADC
∠D = ½◡ABC
◡ADC + ◡ABC = 360°
Let's express the sum of angles:
∠B + ∠D = ½◡ADC + ½◡ABC
Let's put ½ out of brackets:
∠B + ∠D = ½(◡ADC + ◡ABC)
Let's replace the sum of the arcs with their numerical value:
∠B + ∠D = ½ * 360° = 180°
We found that the sum of the opposite angles of an inscribed quadrilateral is 180°. This was what needed to be proven.
The fact that an inscribed quadrilateral has this property (the sum of opposite angles is 180°) does not mean that any quadrilateral whose sum of opposite angles is 180° can be inscribed in a circle. Although in reality this is true. This fact called inscribed quadrilateral test and is formulated as follows: if the sum of the opposite angles of a convex quadrilateral is 180°, then a circle can be described around it (or inscribed in a circle).
You can prove the test for an inscribed quadrilateral by contradiction. Let a quadrilateral ABCD be given whose opposite angles B and D add up to 180°. In this case, angle D does not lie on the circle. Then take a point E on the line containing the segment CD such that it lies on the circle. The result is a cyclic quadrilateral ABCE. This quadrilateral has opposite angles B and E, which means they add up to 180°. This follows from the property of an inscribed quadrilateral.
It turns out that ∠B + ∠D = 180° and ∠B + ∠E = 180°. However, angle D of quadrilateral ABCD with respect to triangle AED is external, and therefore greater than angle E of this triangle. Thus, we have arrived at a contradiction. This means that if the sum of the opposite angles of a quadrilateral adds up to 180°, then it can always be inscribed in a circle.
Theorem 1. The sum of the opposite angles of a cyclic quadrilateral is 180°.
Let a quadrilateral ABCD be inscribed in a circle with center O (Fig. 412). It is required to prove that ∠A + ∠C = 180° and ∠B + ∠D = 180°.
∠A, as inscribed in the circle O, measures 1 / 2 \(\breve(BCD)\).
∠C, as inscribed in the same circle, measures 1 / 2 \(\breve(BAD)\).
Consequently, the sum of angles A and C is measured by the half-sum of arcs BCD and BAD; in sum, these arcs make up a circle, i.e. have 360°.
Hence ∠A + ∠C = 360°: 2 = 180°.
It is similarly proven that ∠B + ∠D = 180°. However, this can be deduced in another way. We know that the sum of the interior angles of a convex quadrilateral is 360°. The sum of angles A and C is equal to 180°, which means that the sum of the other two angles of the quadrilateral also remains 180°.
Theorem 2 (converse). If in a quadrilateral the sum of two opposite angles is equal 180° , then a circle can be described around such a quadrilateral.
Let the sum of the opposite angles of the quadrilateral ABCD be equal to 180°, namely
∠A + ∠C = 180° and ∠B + ∠D = 180° (Fig. 412).
Let us prove that a circle can be described around such a quadrilateral.
Proof. Through any 3 vertices of this quadrilateral you can draw a circle, for example through points A, B and C. Where will point D be located?
Point D can only occupy one of the following three positions: to be inside the circle, to be outside the circle, to be on the circumference of the circle.
Let’s assume that the vertex is inside the circle and takes position D’ (Fig. 413). Then in the quadrilateral ABCD’ we will have:
∠B + ∠D’ = 2 d.
Continuing side AD’ to the intersection with the circle at point E and connecting points E and C, we obtain the cyclic quadrilateral ABCE, in which, by the direct theorem
∠B + ∠E = 2 d.
From these two equalities it follows:
∠D’ = 2 d- ∠B;
∠E = 2 d- ∠B;
but this cannot be, since ∠D’, being external relative to the triangle CD’E, must be greater than angle E. Therefore, point D cannot be inside the circle.
It is also proved that vertex D cannot take position D" outside the circle (Fig. 414).
It remains to recognize that vertex D must lie on the circumference of the circle, i.e., coincide with point E, which means that a circle can be described around the quadrilateral ABCD.
Consequences.
1. A circle can be described around any rectangle.
2. A circle can be described around an isosceles trapezoid.
In both cases, the sum of opposite angles is 180°.
Theorem 3. In the circumscribed quadrilateral, the sums of opposite sides are equal. Let the quadrilateral ABCD be described about a circle (Fig. 415), that is, its sides AB, BC, CD and DA are tangent to this circle.
It is required to prove that AB + CD = AD + BC. Let us denote the points of tangency by the letters M, N, K, P. Based on the properties of tangents drawn to a circle from one point, we have:
Let us add these equalities term by term. We get:
AR + BP + DN + CN = AK + VM + DK + SM,
i.e. AB + CD = AD + BC, which is what needed to be proven.
Other materialsTopic: “Circle described around regular polygon» is discussed in some detail within school curriculum. Despite this, tasks related to this section planimetry cause certain difficulties for many high school students. At the same time, understand the principle of the solution Unified State Exam problems with a circle described around a polygon, graduates with any level of training must.
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