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.
