HSG.C.A.3: Inscribed and Circumscribed Circles and Cyclic Quadrilaterals
In plain English: HSG.C.A.3 is the Common Core geometry standard that asks students to construct the inscribed and circumscribed circles of a triangle and to prove that opposite angles of a quadrilateral inscribed in a circle are supplementary. Angle bisectors locate the center of the inscribed circle and perpendicular bisectors locate the center of the circumscribed circle. It is usually taught in Geometry.
Construct the inscribed and circumscribed circles of a triangle, and prove properties of angles for a quadrilateral inscribed in a circle.
Common Core State Standards for Mathematics · Domain: Circles (C) · Cluster: Understand and apply theorems about circles Also written as HSG-C.A.3 or G-C.3 · Official standard
Students construct the two special circles of a triangle with compass and straightedge. The circumscribed circle passes through all three vertices; its center, the circumcenter, is where the perpendicular bisectors of the sides meet, because every point on a perpendicular bisector is equidistant from the endpoints of that side. The inscribed circle touches all three sides; its center, the incenter, is where the angle bisectors meet, because every point on an angle bisector is equidistant from the sides of the angle.
Students then prove a property of angles for a quadrilateral inscribed in a circle: its opposite angles are supplementary. The proof uses the inscribed angle theorem from HSG.C.A.2, since the two opposite angles intercept arcs that together make the whole circle. Coordinates give students a way to check their constructions exactly.
Learning Objectives
By the end of this lesson, students will be able to:
Construct the circumscribed circle of a triangle with compass and straightedge by locating the intersection of the perpendicular bisectors
Construct the inscribed circle of a triangle by locating the intersection of the angle bisectors and finding the radius along a perpendicular to a side
Explain why each construction works, using the equidistance properties of perpendicular bisectors and angle bisectors
Prove that opposite angles of a quadrilateral inscribed in a circle are supplementary, and use this to find unknown angles
Prior Knowledge Required
Students should already be comfortable with:
Constructing perpendicular bisectors, angle bisectors and perpendicular lines HSG.CO.D.12
Points on a perpendicular bisector are equidistant from the endpoints of the segment HSG.CO.C.9
The inscribed angle theorem: an inscribed angle is half its intercepted arc HSG.C.A.2
Midpoints, slopes of perpendicular lines and the distance formula HSG.GPE.B.5
Show a map with three small towns and a triangular park bordered by three straight roads. Ask two questions side by side:
Warm-Up Prompt
"(1) Where should a cell tower go so that it is the same distance from all three towns? (2) Where should the center of the largest possible circular fountain go inside the park, so that it is the same distance from all three roads? Are these the same kind of point?"
Let students sketch guesses. Guide them to see that question 1 is about distance to points (the vertices), and question 2 is about distance to lines (the sides). Recall the two facts from earlier constructions: points on a perpendicular bisector are equidistant from two points, and points on an angle bisector are equidistant from two lines. Tell students these two facts give the two constructions of the lesson.
Direct Instruction25 minutes
Part 1: Constructions. Demonstrate both constructions under a document camera, using Diagram 1. Students follow along on their own triangle.
Circumscribed circle, step 1: construct the perpendicular bisectors of two sides. With the compass set wider than half the side, draw arcs from both endpoints above and below the side, and join the two crossing points.
Circumscribed circle, step 2: mark the intersection O, the circumcenter. It is on both bisectors, so OA = OB and OB = OC. Put the compass point on O, open it to any vertex and draw the circle; it passes through all three vertices. The third bisector passes through O as well and serves as a check.
Inscribed circle, step 1: construct the bisectors of two angles. From each vertex draw an arc that crosses both sides, then from those two crossing points draw equal arcs that meet inside the angle, and join the vertex to that point.
Inscribed circle, step 2: mark the intersection I, the incenter. It is on both bisectors, so it is the same distance from all three sides.
Inscribed circle, step 3: construct the perpendicular from I to one side and mark its foot F. IF is the radius. Draw the circle with center I through F; it touches each side at exactly one point.
Stress step 5: the radius of the inscribed circle is the perpendicular distance to a side, not the distance to a vertex. Then connect to coordinates, where the constructions can be checked exactly, and to the proof about quadrilaterals.
Part 2: Proof. Let ABCD be inscribed in a circle. Angle A is an inscribed angle that intercepts arc BCD, and angle C intercepts arc DAB. By the inscribed angle theorem, m∠A = (1/2) arc BCD and m∠C = (1/2) arc DAB. The two arcs together make the whole circle, so m∠A + m∠C = (1/2)(360°) = 180°. The same argument gives m∠B + m∠D = 180°. Use Diagram 2 to check the argument with actual numbers.
Circumscribed circle from coordinates
Triangle A(0, 0), B(8, 0), C(2, 6). The perpendicular bisector of AB is x = 4, and the perpendicular bisector of AC passes through (1, 3) with slope -1/3.
Equation: The bisectors meet at O(4, 2); OA = OB = OC = √20 = 2√5
Inscribed circle from coordinates
Triangle B(0, 0), C(14, 0), A(5, 12) has sides 13, 14 and 15. Its angle bisectors meet at I(6, 4).
Equation: The perpendicular from I to BC has foot (6, 0), so r = 4; check: area 84 ÷ semiperimeter 21 = 4
Circumscribed circle of a right triangle
A right triangle has legs 6 and 8. Where is the circumcenter, and what is the radius?
Equation: At the midpoint of the hypotenuse (length 10), so the radius is 5
Opposite angles of an inscribed quadrilateral
ABCD is inscribed in a circle with m∠A = 82° and m∠B = 105°. Find m∠C and m∠D.
In inscribed quadrilateral ABCD, m∠A = (3x + 4)° and m∠C = (2x + 11)°.
Equation: (3x + 4) + (2x + 11) = 180, so x = 33, m∠A = 103° and m∠C = 77°
Guided Practice10-15 minutes
Pairs work three problems and compare with another pair: (a) construct the circumscribed circle of the right triangle with vertices (0, 0), (6, 0) and (0, 4), then confirm by coordinates that the center is the midpoint (3, 2) of the hypotenuse and the radius is √13; (b) a triangle has sides 5, 12 and 13; construct its inscribed circle on a scale drawing and confirm that the radius is 2 (area 30 ÷ semiperimeter 15); (c) quadrilateral ABCD is inscribed with m∠A = 67° and m∠B = 121°, find m∠C (113°) and m∠D (59°). Watch for students who bisect angles when they need perpendicular bisectors, and for students who add adjacent angles instead of opposite ones.
Independent Practice15 minutes
Students work alone: (a) construct the circumscribed circle of the triangle with vertices (1, 1), (7, 1) and (3, 5), and check that the center is (4, 2) with radius √10; (b) construct the inscribed circle of an equilateral triangle with side 12 cm, measure the radius and compare with the exact value 2√3 ≈ 3.46 cm, then explain why the incenter and circumcenter are the same point here; (c) in inscribed quadrilateral PQRS, m∠P = (2y)° and m∠R = (y + 30)°, find y and both angles (y = 50, 100° and 80°).
Closure5 minutes
Exit ticket: "(1) Which lines do you construct to find the center of the circumscribed circle, and which for the inscribed circle? (2) How do you find the radius of the inscribed circle once you have its center? (3) Quadrilateral WXYZ is inscribed in a circle and m∠W = 118°. Find m∠Y." (Answers: perpendicular bisectors of the sides; angle bisectors; the perpendicular distance from the incenter to a side; 62°.)
Differentiation Strategies
For Struggling Students
Provide a two-column card: circumscribed circle, perpendicular bisectors, through the vertices; inscribed circle, angle bisectors, touches the sides
Start the inscribed circle by paper folding: fold each angle so its sides line up, and the creases meet at the incenter
Give a proof frame for the quadrilateral property with the arc names filled in and the reasons left blank
For Advanced Students
Prove that the three perpendicular bisectors of a triangle always meet at one point, and explain where it lies for acute, right and obtuse triangles
Show that the inradius of any triangle equals its area divided by its semiperimeter, by splitting the triangle into three triangles with vertex I
Going further: prove the converse, that a quadrilateral whose opposite angles are supplementary can be inscribed in a circle
Assessment Guidance
What to Look For
For constructions, look for visible compass arcs, not lines drawn by eye, and a check with the third bisector. For the inscribed circle, check that the radius was taken along a perpendicular to a side. For the proof, look for the arcs named correctly (angle A intercepts arc BCD, the arc that does not contain A), the inscribed angle theorem cited as the reason, and the step that the two arcs add up to 360°. A student who only measures four angles and finds they add in pairs to 180° has an example, not a proof.
02
Classroom Activities
3 Activities
1
Place the Cell Tower
20 minPairs
Three towns sit at grid points on a map where each unit is 1 km: Alder at (2, 1), Birch at (12, 1) and Cedar at (4, 7). Pairs construct the circumscribed circle of the triangle to find the point equidistant from all three towns, then check the construction with coordinates.
Procedure
Construct the perpendicular bisectors of two sides with compass and straightedge and mark their intersection
Construct the third bisector as a check: it must pass through the same point
Draw the circle through the three towns and measure its radius with the grid
Check with coordinates: the bisector of Alder-Birch is x = 7, and the tower should be at (7, 8/3), about 5.3 km from each town
Discussion Questions
Why is every point on the perpendicular bisector of Alder-Birch the same distance from those two towns?
If Cedar moved to (7, 0.5), the triangle would be obtuse. Where would the tower go?
Why do you need only two bisectors to find the center?
Modification for Distance Learning
Pairs use a free dynamic geometry tool: plot the three towns, use the perpendicular bisector tool, and drag one town to see how the circumcenter moves inside, onto and outside the triangle.
2
The Largest Fountain in the Plaza
20 minPairs
A triangular plaza has sides of 30 m, 40 m and 50 m. The city wants the largest possible circular fountain inside it. Students construct the inscribed circle on a 1 cm = 5 m scale drawing and find the fountain's radius.
Procedure
Draw the plaza as a 6 cm by 8 cm by 10 cm triangle
Construct the bisectors of two angles and mark the incenter; construct the third as a check
Construct the perpendicular from the incenter to one side, and set the compass to that length to draw the circle
Measure the radius (2 cm on the drawing, so 10 m in the plaza) and confirm: area 600 m² ÷ semiperimeter 60 m = 10 m
Discussion Questions
Why would a fountain centered at the circumcenter not fit inside the plaza?
A student used the distance from the incenter to a corner as the radius. What would happen?
Does the fountain touch each side at the midpoint of that side?
Paper Folding Variation
Cut out the triangle and fold each angle so its two sides lie on top of each other. The three creases are the angle bisectors and meet at the incenter.
3
Inscribed Quadrilateral Proof
20 minGroups of 3-4
Groups collect data from inscribed quadrilaterals, write the proof that opposite angles are supplementary, and test which familiar quadrilaterals can be inscribed in a circle.
Procedure
Each student draws a circle and four points on it, joins them in order and measures all four angles; the group records the sums of opposite angles
The group writes a proof: name the arc each angle intercepts, apply the inscribed angle theorem and add the arcs
Use the property to test figures: can a rectangle, a non-square rhombus or a parallelogram with a 70° angle be inscribed? (Only the rectangle: the others have opposite angles that are equal but not 90°, so they cannot add to 180°.)
Discussion Questions
Why do the two intercepted arcs always add up to 360°?
What must be true of a parallelogram that is inscribed in a circle?
Does the proof need the center of the circle to be inside the quadrilateral?
Challenge Variation
Going further: groups investigate the converse with dynamic geometry. Draw a quadrilateral whose opposite angles are supplementary, construct the circle through three of its vertices, and check whether the fourth vertex lies on it.
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Diagrams & Visual Aids
2 diagrams
Diagram 1: Constructing the Circumscribed and Inscribed Circles
Left: the perpendicular bisectors of the sides of triangle A(0, 0), B(8, 0), C(2, 6) meet at O(4, 2), which is 2√5 from every vertex, so the circle through the vertices has center O. Right: the angle bisectors of triangle B(0, 0), C(14, 0), A(5, 12) meet at I(6, 4). The perpendicular from I to BC has length 4, the radius of the inscribed circle, which touches each side once. Both triangles are drawn to scale.
Diagram 2: Opposite Angles of an Inscribed Quadrilateral
ABCD is inscribed in a circle with arcs of 60°, 110°, 94° and 96°. Each angle is half the arc it intercepts, so m∠A = 102°, m∠B = 95°, m∠C = 78° and m∠D = 85°. Opposite angles intercept arcs that together make the whole 360° circle, so each pair adds to 180°. The vertices are placed exactly from the arc measures.
04
Homework Assignment
~30 min
HSG.C.A.3 Homework: Inscribed and Circumscribed Circles
Directions: Do each construction with compass and straightedge on graph paper and leave every arc visible. Then check it with coordinates or a calculation. Give a reason for every step of a proof.
Part 1: Constructions (Problems 1-3)
Plot P(0, 0), Q(10, 0) and R(4, 8). Construct the circumscribed circle of triangle PQR. Then find the exact coordinates of its center and its radius, and show that the center is the same distance from all three vertices.
Plot D(0, 0), E(12, 0) and F(0, 9). Construct the inscribed circle of triangle DEF. Measure its radius, then confirm it by dividing the area of the triangle by its semiperimeter, and give the coordinates of the center.
Plot J(0, 0), K(10, 0) and L(2, 3). Construct the circumscribed circle of triangle JKL. Find the exact center, state whether it is inside, on or outside the triangle, and explain what property of the triangle causes this.
Part 2: Inscribed Quadrilaterals (Problems 4-5)
Quadrilateral JKLM is inscribed in a circle with m∠J = 94° and m∠K = 71°. Find m∠L and m∠M.
Quadrilateral ABCD is inscribed in a circle with m∠A = (4x - 6)°, m∠B = (2x + 13)° and m∠C = (x + 21)°. Find x and all four angles.
Part 3: Proof (Problem 6)
Write a proof that opposite angles of a quadrilateral inscribed in a circle are supplementary. Then use the result to prove that a parallelogram inscribed in a circle must be a rectangle.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Constructions
Correct bisectors with visible arcs; circle through vertices or touching sides
Correct lines but circle or radius inaccurate
Wrong lines used or no construction
Coordinate Check
Exact center and radius, with equal distances shown
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
To construct the circumscribed circle of a triangle, which lines do you construct to find its center?
Answer: C
Points on the perpendicular bisector of a side are equidistant from its endpoints, so the intersection of the bisectors is equidistant from all three vertices. Choice A finds the center of the inscribed circle instead. Choice B finds the centroid, which is usually not equidistant from the vertices.
Question 2 of 20 · Multiple Choice
To construct the inscribed circle of a triangle, which lines do you construct to find its center?
Answer: A
Points on an angle bisector are equidistant from the two sides of the angle, so the intersection of the angle bisectors is equidistant from all three sides. Choice B finds the circumcenter, which is equidistant from the vertices, not the sides.
Question 3 of 20 · Multiple Choice
After finding the incenter I of a triangle, how do you set the compass to draw the inscribed circle?
Answer: B
The radius of the inscribed circle is the perpendicular distance from I to a side. Choice A gives a circle that passes outside the triangle, a common error. Choice D works only when the circle happens to touch that side at its midpoint, as in an equilateral triangle.
Question 4 of 20 · Multiple Choice
Where is the circumcenter of a right triangle?
Answer: D
The right angle is inscribed in the circumscribed circle, so the hypotenuse is a diameter and its midpoint is the center. Choice A is where the legs meet, and that vertex is not the same distance from all three vertices. Choice C describes an obtuse triangle.
Question 5 of 20 · Multiple Choice
What is the center of the circumscribed circle of the triangle with vertices (0, 0), (6, 0) and (0, 8)?
Answer: A
The triangle has a right angle at (0, 0), so the center is the midpoint of the hypotenuse from (6, 0) to (0, 8): (3, 4), with radius 5. Choice B is the centroid, the average of the vertices. Choice C is the incenter. Choice D is the fourth corner of the rectangle.
Question 6 of 20 · Multiple Choice
A right triangle has legs 20 and 21 and hypotenuse 29. What is the radius of its inscribed circle?
Answer: C
Area = (1/2)(20)(21) = 210 and semiperimeter = (20 + 21 + 29)/2 = 35, so r = 210 ÷ 35 = 6. Choice B is half the hypotenuse, which is the radius of the circumscribed circle. Choice A is half of a leg.
Question 7 of 20 · Multiple Choice
Quadrilateral ABCD is inscribed in a circle and m∠A = 72°. What is m∠C?
Answer: B
Opposite angles of an inscribed quadrilateral are supplementary: 180° - 72° = 108°. Choice A assumes opposite angles are equal, which is true in a parallelogram but not here. Choice D subtracts from 90° instead of 180°.
Question 8 of 20 · Multiple Choice
Quadrilateral ABCD is inscribed in a circle and m∠B = 97°. What is m∠D?
Answer: D
B and D are opposite angles, so m∠D = 180° - 97° = 83°. Choice B subtracts from 360°, the sum of all four angles, instead of from 180°. Choice C halves the angle as if it were an arc.
Question 9 of 20 · Multiple Choice
Which kind of quadrilateral can always be inscribed in a circle?
Answer: B
A rectangle has four 90° angles, so each pair of opposite angles adds to 180°, and its diagonals are congruent diameters of a circle through all four vertices. A rhombus or parallelogram with a 70° angle has opposite angles 70° and 70°, which cannot be supplementary, so choices A and C fail. Many kites cannot be inscribed either.
Question 10 of 20 · Multiple Choice
In inscribed quadrilateral ABCD, m∠A = (5x + 10)° and m∠C = (3x + 10)°. What is x?
Answer: C
(5x + 10) + (3x + 10) = 180, so 8x = 160 and x = 20, giving 110° and 70°. Choice D sets the sum equal to 360°. Choice B ignores the two constant terms and solves 8x = 180.
Question 11 of 20 · Multiple Choice
In the proof that opposite angles A and C of inscribed quadrilateral ABCD are supplementary, which reason is correct?
Answer: A
m∠A = (1/2) arc BCD and m∠C = (1/2) arc DAB, and these arcs cover the circle, so the angles add to 180°. Choice B is false for most quadrilaterals; the property depends on the circle. Choices C and D are true only for special quadrilaterals.
Question 12 of 20 · Multiple Choice
Where is the circumcenter of an obtuse triangle?
Answer: D
The obtuse angle is an inscribed angle greater than 90°, so the arc it intercepts is greater than 180°. The center of the circle is then on the far side of the longest side from the obtuse vertex, outside the triangle. Choice B is the right-triangle case. Choice A holds for acute triangles.
Question 13 of 20 · Multiple Choice
What is the center of the circumscribed circle of the triangle with vertices (0, 0), (8, 0) and (4, 8)?
Answer: B
The perpendicular bisector of the base is x = 4. The bisector of the side from (0, 0) to (4, 8) passes through (2, 4) with slope -1/2, so at x = 4, y = 4 - 1 = 3. The center (4, 3) is 5 units from each vertex. Choice C is the centroid. Choice A is the midpoint of the altitude.
Question 14 of 20 · Multiple Choice
A park district wants a first-aid station that is the same distance from three schools. Which point of the triangle formed by the schools should it use?
Answer: C
The circumcenter is equidistant from the three vertices, here the schools. Choice A is equidistant from the three sides, so it would suit a point the same distance from three straight roads. Choice D works only when the triangle has a right angle.
Question 15 of 20 · Short Answer
A triangle has vertices (0, 0), (15, 0) and (0, 20). Find the radius and the center of its inscribed circle.
The hypotenuse is √(15² + 20²) = 25. Area = (1/2)(15)(20) = 150 and semiperimeter = (15 + 20 + 25)/2 = 30, so r = 150 ÷ 30 = 5. The circle touches both legs, which lie on the axes, so its center is 5 units from each: (5, 5).
Question 16 of 20 · Short Answer
Find the center and radius of the circumscribed circle of the triangle with vertices A(0, 0), B(10, 0) and C(6, 6).
The perpendicular bisector of AB is x = 5. The bisector of AC passes through (3, 3) with slope -1, so y = -x + 6, and at x = 5, y = 1. The center is (5, 1) and the radius is √(5² + 1²) = √26. Check: the distance to C is √(1² + 5²) = √26.
Question 17 of 20 · Short Answer
Quadrilateral ABCD is inscribed in a circle with m∠A = 88° and m∠B = 99°. Find m∠C and m∠D.
Write a paragraph proof that opposite angles B and D of an inscribed quadrilateral ABCD are supplementary.
Angle B is an inscribed angle that intercepts arc CDA, and angle D intercepts arc ABC. By the inscribed angle theorem, m∠B = (1/2) arc CDA and m∠D = (1/2) arc ABC. The two arcs together make the whole circle, so their measures add to 360°. Therefore m∠B + m∠D = (1/2)(360°) = 180°, and the angles are supplementary.
Question 19 of 20 · Short Answer
Explain why the point where the angle bisectors of a triangle meet is the center of a circle that touches all three sides.
Every point on the bisector of an angle is the same distance from the two sides of that angle. The intersection point I is on the bisectors of angles A and B, so it is the same distance from sides AB and AC and from sides AB and BC, so it is equidistant from all three sides. A circle with center I and radius equal to that perpendicular distance touches each side at the foot of the perpendicular.
Question 20 of 20 · Short Answer
A quadrilateral has angles of 110°, 75°, 80° and 95°, in order around the figure. Can it be inscribed in a circle? Explain.
No. If it were inscribed, its opposite angles would be supplementary. But the first and third angles add to 110° + 80° = 190°, not 180° (and the others add to 75° + 95° = 170°), so no circle can pass through all four vertices.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSG.C.A.3 mean?
HSG.C.A.3 asks students to do two things: construct the inscribed and circumscribed circles of a triangle, and prove properties of the angles of a quadrilateral inscribed in a circle. The main property is that opposite angles of an inscribed quadrilateral are supplementary.
Is HSG.C.A.3 taught in Geometry or Algebra 2?
It is usually taught in high school Geometry, in the circles unit, after the basic constructions (HSG.CO.D.12) and the inscribed angle theorem (HSG.C.A.2).
What is the difference between the incenter and the circumcenter?
The incenter is the center of the inscribed circle: it is where the angle bisectors meet, and it is the same distance from all three sides. The circumcenter is the center of the circumscribed circle: it is where the perpendicular bisectors of the sides meet, and it is the same distance from all three vertices.
Why do perpendicular bisectors locate the center of the circumscribed circle?
Every point on the perpendicular bisector of a segment is equidistant from the segment's endpoints. The point on two bisectors is equidistant from all three vertices, so a circle centered there passes through all three. That same point must also lie on the third bisector.
How do you find the radius of the inscribed circle?
Construct the perpendicular from the incenter to any side; the length of that segment is the radius. Measuring from the incenter to a vertex is a common error and gives a circle that crosses the sides. As a numerical check, the radius equals the area of the triangle divided by its semiperimeter.
Where is the circumcenter of acute, right and obtuse triangles?
It is inside an acute triangle, at the midpoint of the hypotenuse of a right triangle, and outside an obtuse triangle, beyond the longest side. The incenter, by contrast, is always inside the triangle.
Why are opposite angles of an inscribed quadrilateral supplementary?
Each of the two opposite angles is an inscribed angle, so each is half of the arc it intercepts. Those two arcs together make the whole circle, 360°, so the angles add to half of 360°, which is 180°.
Can every quadrilateral be inscribed in a circle?
No. A quadrilateral can be inscribed only if its opposite angles are supplementary. A rhombus or parallelogram with a 70° angle cannot be inscribed, while every rectangle and every isosceles trapezoid can. Proving that supplementary opposite angles guarantee a circle (the converse) goes beyond this standard.
What tools can students use for these constructions?
Compass and straightedge are the classic tools. Paper folding works well for angle bisectors and perpendicular bisectors, and dynamic geometry software lets students drag vertices and watch the centers move. Whatever the tool, students should explain why the construction works.
What mistakes do students make on HSG.C.A.3?
Common errors include:
using angle bisectors for the circumscribed circle or perpendicular bisectors for the inscribed circle
setting the inscribed circle's radius to the distance from the incenter to a vertex
adding adjacent angles of an inscribed quadrilateral instead of opposite ones
checking four measured angles and calling it a proof
07
Related Standards
5 standards
These standards connect to HSG.C.A.3: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSG.CO.D.12Prerequisite
Make formal constructions, including bisecting angles and perpendicular bisectors
