Cyclic Quadrilateral Cyclic Quadrilateral In Euclidean geometry, a cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic. The center of the circle and its radius are called the circumcenter and the circumradius respectively. Other names for these quadrilaterals are concyclic quadrilateral and chordal quadrilateral, the latter since the sides of the quadrilateral are chords of the circumcircle. Usually the quadrilateral is assumed to be convex, but there are also crossed cyclic quadrilaterals. The formulas and properties given below are valid in the convex case. The word cyclic is from the Greek kuklos which means "circle" or "wheel".
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All triangles have a circumcircle, but not all quadrilaterals do. An example of a quadrilateral that cannot be cyclic is a non-square rhombus. The section characterizations below states what necessary and sufficient conditions a quadrilateral must satisfy to have a circumcircle. Theorem 1: Opposite Angles of a Cyclic Quadrilateral will add up to 1800 (Supplementary to One Another), i.e., their sum would equals to two right angles. Let’s consider a cyclic quad ABCD. Join the diagonals AC and BD. Proof: As we can see here, 3 angles in any triangle would sum up to two 900 angles, in △ABC, ∠ABC + ∠BCA + ∠CAB = Two Right Angles eqn (1). In a circle inscribed angles having common circumference and base are congruent to each other. Thus, ∠CAB = ∠BDC and ∠BCA = ∠ADB. Therefore, we can write: ∠ADC = ∠ADB + ∠BDC =∠BCA + ∠CAB eqn (2) By using eqn (1), we have ∠BCA + ∠CAB = Two Right Angles−∠ABC eqn (3), Equating equation (2) and equation (3), we get
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∠ADC = Two Right Angles − ∠ABC, or ∠ABC + ∠ADC = Two Right Angles. Similarly, we can also prove that ∠BAC + ∠BCD=Two Right Angles. Hence the theorem is proved. Theorem 2: If opposite angles of a quadrilateral are supplementary, we may circumscribe a circle about the Quadrilateral. Let’s consider a quadrilateral ABCD in which opposite angles are supplementary. That is, ∠A + ∠C = ∠B + ∠D = 180∘equation (1) Proof: We are actually going to prove this quadrilateral cyclic here. For this, let's say that circle does not pass through the vertex D of quadrilateral, but cuts line AD or AD′ when it is produced, to D′. This proves that quad ABCD′ as a cyclic quadrilateral. Hence, opposite angles are supplementary; that is, ∠B + ∠D′ = ∠A + ∠C = 180∘equation (2), From equation (1) and equation (2), we can write: ∠B + ∠D = ∠B + ∠D′, or ∠D = ∠D', This contradicts as points D and D′ are distinct points and thus do not coincide. So, our postulation is wrong that circle passing through A, B, and C do not pass through D. Therefore, ABCD is a cyclic quadrilateral. Hence theorem is proved.
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