Construct the inscribed and circumscribed circles of a triangle and prove that opposite angles of an inscribed quadrilateral are supplementary

Learning goal: Construct the inscribed circle and the circumscribed circle of a triangle with compass and straightedge, explain why each construction works, and prove and apply the property that opposite angles are supplementary for a quadrilateral inscribed in a circle.

Two circles that belong to every triangle. The circumscribed circle (circumcircle) of a triangle passes through all three vertices; its center is the circumcenter and its radius is usually written R. The inscribed circle (incircle) lies inside the triangle and is tangent to all three sides; its center is the incenter and its radius is written r. Every triangle has exactly one of each. The constructions rest on two locus facts from earlier work: the set of points equidistant from two points A and B is the perpendicular bisector of segment AB, and the set of points inside an angle equidistant from its two sides is the angle bisector.

Constructing the circumscribed circle. Given triangle ABC, construct the perpendicular bisector of AB (arcs of equal radius from A and from B, meeting above and below; join the meeting points) and the perpendicular bisector of BC. They meet at a point O. Because O is on the perpendicular bisector of AB, OA = OB; because O is on the perpendicular bisector of BC, OB = OC. So OA = OB = OC, and O is also on the perpendicular bisector of AC without our drawing it. The circle centered at O with radius OA passes through all three vertices. The three perpendicular bisectors are concurrent, and the point where they meet is the circumcenter. It lies inside an acute triangle, on the hypotenuse (at its midpoint) for a right triangle, and outside an obtuse triangle.

Constructing the inscribed circle. Construct the bisectors of angles A and B (an arc across the angle from the vertex, then equal arcs from the two crossing points; join the vertex to where they meet). They meet at a point I inside the triangle. A point on the bisector of angle A is equidistant from sides AB and AC; a point on the bisector of angle B is equidistant from sides BA and BC. So I is at the same distance r from all three sides, and it lies on the bisector of angle C as well. To find r, drop a perpendicular from I to any side (construct the perpendicular from a point to a line). The circle centered at I with radius r touches each side at the foot of that perpendicular, so it is tangent to all three sides: the radius to the touching point is perpendicular to the side, which is the tangent condition from the previous lesson. The incenter is always inside the triangle.

A worked model with coordinates. Take the right triangle with vertices A(0, 0), B(8, 0) and C(0, 6); its legs are 8 and 6 and its hypotenuse BC = √(64 + 36) = 10. The perpendicular bisector of AB (a horizontal segment from x = 0 to x = 8) is the vertical line x = 4; the perpendicular bisector of AC is the horizontal line y = 3. They meet at O(4, 3), the circumcenter. Check all three distances: OA = √(16 + 9) = 5, OB = √((8 − 4)² + 3²) = 5, OC = √(4² + (6 − 3)²) = 5. So R = 5, and O(4, 3) is the midpoint of the hypotenuse BC, exactly as expected for a right triangle. For the incircle, the bisector of the right angle at A is the line y = x, and the incenter has coordinates (r, r) where r is its distance from both legs. Its distance from the hypotenuse, the line 3x + 4y = 24, must also be r. For a right triangle with legs a, b and hypotenuse c, the inradius is r = (a + b − c)/2 = (8 + 6 − 10)/2 = 2, so the incenter is I(2, 2). Check with the distance-to-a-line formula: |3·2 + 4·2 − 24|/√(3² + 4²) = |−10|/5 = 2. A second check: the area of any triangle equals r times half its perimeter; here the area is ½·8·6 = 24, the half-perimeter is (8 + 6 + 10)/2 = 12, and 24/12 = 2.

The cyclic quadrilateral theorem. A quadrilateral is inscribed in a circle (also called cyclic) when all four vertices lie on the circle. Theorem: opposite angles of an inscribed quadrilateral are supplementary. Proof: let ABCD be inscribed in a circle. Angle A is an inscribed angle whose sides pass through B and D, so it intercepts arc BCD, the arc that does not contain A; by the inscribed angle theorem, angle A = ½·arc BCD. Likewise angle C intercepts arc DAB, so angle C = ½·arc DAB. The two arcs BCD and DAB together make up the whole circle, so arc BCD + arc DAB = 360°. Therefore angle A + angle C = ½(arc BCD + arc DAB) = ½·360° = 180°. The same argument with arcs ABC and CDA shows angle B + angle D = 180°.

Applying the theorem. If an inscribed quadrilateral has angle A = 70° and angle B = 95°, then angle C = 180° − 70° = 110° and angle D = 180° − 95° = 85°. Check the angle sum of a quadrilateral: 70 + 95 + 110 + 85 = 360°. If angle A = (3x + 10)° and angle C = (2x − 5)°, then 3x + 10 + 2x − 5 = 180, so 5x = 175 and x = 35, giving angle A = 115° and angle C = 65°. A useful consequence: if a parallelogram is inscribed in a circle, its opposite angles are equal (parallelogram) and supplementary (cyclic), so each is 90° and the parallelogram is a rectangle. The converse of the theorem also holds for convex quadrilaterals, and is used as a test: if one pair of opposite angles of a convex quadrilateral is supplementary, a circle passes through all four vertices.

The misconception: any quadrilateral has a circumscribed circle. Three non-collinear points always lie on a circle, which is why every triangle has a circumcircle. A fourth point usually does not lie on that circle; a general quadrilateral has no circumscribed circle at all. A rhombus that is not a square, for example, has opposite angles equal but not supplementary, so it cannot be inscribed in a circle. The theorem's condition, "inscribed in a circle," is what makes the conclusion available.

A second misconception: the incenter and circumcenter are the same point. They coincide only for an equilateral triangle. In the 6-8-10 triangle the circumcenter is (4, 3) and the incenter is (2, 2), two different points with two different radii, 5 and 2. Remembering which construction uses perpendicular bisectors (circumscribed, through the vertices) and which uses angle bisectors (inscribed, tangent to the sides) keeps them apart.

Worked circumscribed circle: Triangle with vertices P(−2, 0), Q(6, 0) and S(2, 8). The perpendicular bisector of PQ is the vertical line x = 2 (midpoint (2, 0)). The perpendicular bisector of PS: midpoint is (0, 4), slope of PS is 8/4 = 2, so the bisector has slope −1/2 and equation y − 4 = −½(x − 0), that is y = −½x + 4. At x = 2, y = 3. Circumcenter O(2, 3). Check: OP = √(16 + 9) = 5, OQ = √(16 + 9) = 5, OS = √(0 + 25) = 5. Radius R = 5.

Worked inscribed circle: For the 8-6-10 right triangle A(0, 0), B(8, 0), C(0, 6), the inradius is r = (8 + 6 − 10)/2 = 2 and the incenter is I(2, 2). The points of tangency are the feet of the perpendiculars: on AB it is (2, 0), on AC it is (0, 2), and on the hypotenuse 3x + 4y = 24 it is the foot from (2, 2), which is (2, 2) + 2·(3/5, 4/5) = (3.2, 3.6). Check that (3.2, 3.6) is on the hypotenuse: 3(3.2) + 4(3.6) = 9.6 + 14.4 = 24. Each tangency point is at distance 2 from I, and the radius there is perpendicular to the side.

Worked cyclic quadrilateral proof: ABCD is inscribed in a circle. Angle B intercepts arc CDA and angle D intercepts arc ABC. These two arcs share no interior points and together cover the circle, so arc CDA + arc ABC = 360°. By the inscribed angle theorem angle B = ½·arc CDA and angle D = ½·arc ABC, so angle B + angle D = ½·360° = 180°.

Worked application: In an inscribed quadrilateral, angle A = (4x − 8)°, angle C = (2x + 20)° and angle B = 3x°. From A + C = 180: 6x + 12 = 180, x = 28, so A = 104°, C = 76°, B = 84°, and D = 180° − 84° = 96°. Check: 104 + 84 + 76 + 96 = 360.

Construction record: On one sheet, construct the circumscribed circle of an acute triangle and on another the inscribed circle of the same triangle, leaving all arcs visible. Beside each, write the locus fact that justifies the construction (perpendicular bisector: equidistant from two points; angle bisector: equidistant from two lines), name the center, and measure the radius. Then repeat the circumcircle construction on an obtuse triangle and record where its circumcenter falls.

Independent performance: Write a proof that the opposite angles of an inscribed quadrilateral are supplementary, then use the theorem to solve a problem you design in which all four angles are given as expressions in x. Finally explain, with a specific example such as a non-square rhombus, why a quadrilateral that is not inscribed in a circle need not satisfy the theorem. Success criteria: the proof names the arcs each angle intercepts and states that they sum to 360°; the algebra is solved and all four angles are checked to total 360°; the counterexample gives actual angle measures that are not supplementary; and the write-up distinguishes the inscribed circle (tangent to sides, angle bisectors) from the circumscribed circle (through vertices, perpendicular bisectors).