A) construct a circle of any radius, and draw a chord on it. then construct the radius of the circle that bisects the chord. measure the angle between the chord and the radius. what can you conclude about the intersection of a chord and the radius that bisects it? take a screenshot of your construction, save it, and insert the image below your answer. b) write a paragraph proof of your conclusion in part a. to begin your proof, draw radii oa and oc. c) draw a circle of any radius, and draw a chord on it. construct the radius of the circle that is perpendicular to the chord. measure the line segments into which the radius divides the chords. how are the line segments related? what can you conclude about the intersection of a chord and a radius that is perpendicular to it? take a screenshot of your construction, save it, and insert the image below your answer. d) write a paragraph proof of your conclusion in part c. to begin your proof, draw radii oa and oc.

Diagram for part A and part C is attached herewith.

Solving for part B:-

Given is the circle O. We start with drawing two radii OA and OC, then we join two points A and C to make a chord AC of the circle. Now The radius of the circle intersects the chord AC at point B such that it bisects AC into two equal parts AB and BC. Now we have two triangles ΔOBA and ΔOBC. In these two triangles, we have OA=OC (radii of circle), OB=OB (reflexive property), and BA=BC (given in the question). Using SSS congruency of triangles we can say ΔOBA≡ΔOBC and using CPCTC, we can conclude ∡OBA=∡OBC (=90°). Hence OB⊥AC i.e. OB is perpendicular to AC.

Solving for part D:-

Given is the circle O. We start with drawing two radii OA and OC, then we join two points A and C to make a chord AC of the circle. Now The radius of the circle intersects the chord AC at point B such that AB is perpendicular to AC i.e. ∡B=90°. Now we have two Right triangles ΔOBA and ΔOBC. In these two triangles, we have OA=OC (radii of circle), OB=OB (reflexive property). Using HL congruency of right triangles we can say ΔOBA≡ΔOBC and using CPCTC, we can conclude BA=BC. Hence OB bisects AC into AB=BC.

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step-by-step explanation:

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