Essential Question: How do I construct inscribed circles, circumscribed circles Standard: MCC9-12.G.C.3 & 4

1) Construct the perpendicular bisectors of the sides of the triangle and label the point of intersection F. From point A construct arcs 3 & 4 and draw a line connecting the intersections of the arcs From point B construct arcs 7 & 8 and draw a line connecting the intersections of the arcs Construction #1 Given a triangle construct the circumscribed circle. Given: Triangle ABC PROCEDURE: A B C Bisect segment AC; Using a radius greater than 1/2AC from point C construct arcs 1 & Bisect segment BC; Using a radius greater than 1/2BC from point C construct arcs 5 & F 2) Set your compass pointer to point F and the radius to measure FC. 3) Draw the circle with center F, that passes through the vertices A, B, & C r a d i u s Now construct the perpendicular bisector of segment AB and label point F, where the 3 perpendicular bisectors meet.

Making Connections (Construction #1) Point F is the circumcenter of the triangle, because it is the center of the circle that circumscribes the triangle. It is equidistant to each of the vertices of the triangle. The edges of the triangle ABC are now chord AB, chord BC, and cord AC of circle.  BFC,  AFB, and  CFA are central angles of circle F. They are congruent to the intercepted arcs.  BAC,  ACB, and  CBA are inscribed angles of circle F. They are half the measure of the intercepted arcs. A B C F

F Construction #2 Given a triangle construct the inscribed circle. Given: Triangle ABC PROCEDURE: A B C 1) Construct the angle bisectors of angles A, B, & C, to get a point of intersection and call it F. 2) Construct a perpendicular to side AC from point F, and label this point G. 3) Put your pointer on point F and set your radius to FG. 4) Draw the circle using F as the center and it should be tangent to all the sides of the triangle. G XY

Making Connections (Construction #2) Point F is the incenter of the triangle, because it is the center of the circle that inscribes the triangle. It is equidistant to each of the sides of the triangle. The edges of the triangle ABC are now tangent AB, tangent BC, and tangent AC of circle F. A B C F  SFT,  TFU, and  UFS are central angles of circle F. They are congruent to the intercepted arcs. S U T