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 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 (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 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;
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. 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), 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.
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INSCRIBED AND CIRCULAR POLYGONS,
§ 106. PROPERTIES OF INSCRIBED AND DESCRIBED QUADRIAGONS.
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 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 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(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 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;
/ D" = / E,
but this cannot be, because / 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 a 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 (§ 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 a cyclic quadrilateral there are two opposite angles ratio is 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 circle is said to be inscribed in a quadrilateral if all sides of the quadrilateral are tangent to the circle.
The center of this circle is the intersection point of the bisectors of the corners of the quadrilateral. In this case, the radii drawn to the tangent points are perpendicular to the sides of the quadrilateral
A circle is called circumscribed about a quadrilateral if it passes through all its vertices.
The center of this circle is the point of intersection of the perpendicular bisectors to the sides of the quadrilateral
Not every quadrilateral can be inscribed with a circle, and not every quadrilateral can be inscribed with a circle.
PROPERTIES OF INSCRIBED AND CIRCULAR Quadrilaterals
THEOREM In a convex inscribed quadrilateral, the sums of opposite angles are equal to each other and equal to 180°.
THEOREM Conversely: if in a quadrilateral the sums of opposite angles are equal, then a circle can be described around the quadrilateral. Its center is the point of intersection of the perpendicular bisectors to the sides.
THEOREM If a circle is inscribed in a quadrilateral, then the sums opposing sides its equal.
THEOREM Conversely: if in a quadrilateral the sums of opposite sides are equal, then a circle can be inscribed in it. Its center is the point of intersection of the bisectors.
Corollaries: of all parallelograms, only around a rectangle (in particular, around a square) can a circle be described.
Of all the parallelograms, only a rhombus (in particular a square) can inscribe a circle (the center is the point of intersection of the diagonals, the radius is equal to half height).
If a circle can be described around a trapezoid, then it is isosceles. A circle can be described around any isosceles trapezoid.
If a circle is inscribed in a trapezoid, then its radius is equal to half the height.
Tasks with solutions
1. Find the diagonal of a rectangle inscribed in a circle whose radius is 5.
The center of a circle circumscribed about a rectangle is the point of intersection of its diagonals. Therefore, the diagonal AC equals 2 R. That is AC=10
Answer: 10.
2. A circle is described around a trapezoid, the bases of which are 6 cm and 8 cm, and the height is 7 cm. Find the area of this circle.
Let DC=6, AB=8. Since a circle is circumscribed around a trapezoid, it is isosceles.
Let's draw two heights DM and CN.Since the trapezoid is isosceles, then AM=NB=
Then AN=6+1=7
From a triangle ANS using the Pythagorean theorem we find AC.
From a triangle CВN using the Pythagorean theorem we find Sun.
The circumscribed circle of a trapezoid is also the circumscribed circle of a triangle. DIA
Let's find the area of this triangle in two ways using the formulas
Where h- height and - base of triangle
Where R is the radius of the circumscribed circle.
From these expressions we obtain the equation. Where
The area of the circle will be equal to
3. Angles and quadrilaterals are related as . Find the angle if a circle can be described around a given quadrilateral. Give your answer in degrees
It follows from the condition that .Since a circle can be described around a quadrilateral, then
We get the equation 
. Then . The sum of all the angles of a quadrilateral is 360º. Then
. where do we get that
4.The sides of a trapezoid circumscribed about a circle are 3 and 5. Find the midline of the trapezoid.
Then middle line equal to 
5. Perimeter rectangular trapezoid circumscribed about a circle is 22, its major side is equal to 7. Find the radius of the circle.
In a trapezoid, the radius of the inscribed circle is equal to half the height. Let's draw the height of the SC.
Then 
.
Since a circle is inscribed in a trapezoid, the sums of the lengths of opposite sides are equal. Then
Then the perimeter
We get the equation
6. The bases of an isosceles trapezoid are 8 and 6. The radius of the circumscribed circle is 5. Find the height of the trapezoid.
Let O be the center of the circle circumscribed about the trapezoid. Then .
Let's draw the height KH through point O
Then 
, where KO and OH are heights and at the same time medians isosceles triangles DOC and AOB. Then 
According to the Pythagorean theorem.