AMSCO'S Geometry. New York - Rye High School

Hands-On ActivityConsider any regular polygon. Construct the angle bisectors of eachinterior angle. Since the interior angles are all congruent, the anglesformed are all congruent. Since the sides of the regular polygon areall congruent, congruent isosceles triangles are formed by ASA. Anytwo adjacent triangles share a common leg. Therefore, they all sharethe same vertex. Since the legs of the triangles formed are all congruent,the vertex is equidistant from the vertices of the regular polygon.This common vertex is the center of the regular polygon.Angles Formed by Tangents, Chords, and Secants 567In this Hands-On Activity, we will use the center of a regular polygon to inscribe a circle inthe polygon.a. Using geometry software or compass, protractor, and straightedge, construct a square, a regularpentagon, and a regular hexagon. For each figure:(1) Construct the center P of the regular polygon. (The center is the intersection of the anglebisectors of a regular polygon.)(2) Construct an apothem or perpendicular from P to one of the sides of regular polygon.(3) Construct a circle with center P and radius equal to the length of the apothem.b. Prove that the circles constructed in part a are inscribed inside of the polygon. Prove:(1) The apothems of each polygon are all congruent.(2) The foot of each apothem is on the circle.(3) The sides of the regular polygon are tangent to the circle.c. Let r be the distance from the center to a vertex of the regular polygon. Since the center isequidistant from each vertex, it is possible to circumscribe a circle about the polygon withradius r. Let a be the length of an apothem and s be the length of a side of the regular polygon.How is the radius, r, of the circumscribed circle related to the radius, a, of the inscribed circle?13-5 ANGLES FORMED BY TANGENTS, CHORDS, AND SECANTSAngles Formed by a Tangent and a ChordIn the diagram, AB gis tangent to circle O at A, AD isa chord, and AC is a diameter. When CD is drawn,ADC is a right angle because it is an angle inscribed ina semicircle, and ACD is the complement of CAD.Also, CA ' AB, BAC is a right angle, and DAB is thecomplement of CAD. Therefore, since complements ofthe same angle are congruent, ACD DAB.We1can conclude that since mACD = X , then1mDAB = X2 mAD 2 mAD.COADB