In figure 3.98, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

Question

In figure 3.98, seg AB is a diameter of a circle with centre O. The bisector of ∠ ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks. Figure: C and D lie on opposite sides of diameter AB. Proof: Draw seg OD. ∠ ACB = □ ………. angle inscribed in semicircle ∠ DCB = □ ………. CD is the bisector of ∠ C m(arc DB) = □ ………. inscribed angle theorem ∠ DOB = □ ………. definition of measure of an arc (I) seg OA ≅ seg OB ………. □ (II) therefore line OD is □ of seg AB ………. From (I) and (II) therefore seg AD ≅ seg BD

Doubtly AI · Answer

Answer: OD is the perpendicular bisector of AB, so D is equidistant from A and B. Draw seg OD. ∠ ACB = 90^° …… angle inscribed in semicircle ∠ DCB = 45^° …… CD is the bisector of ∠ C m(arc DB) = 90^° …… inscribed angle theorem ( 2 × 45^° ) ∠ DOB = 90^° …… definition of measure of an arc (I) seg OA ≅ seg OB …… radii of the same circle (II) therefore line OD is the perpendicular bisector of seg AB …… from (I) and (II) Every point on the perpendicular bisector of a segment is equidistant from its end points, therefore seg AD ≅ seg BD.

In figure 3.98, seg AB is a diameter of a circle with centre O. The bisector of ∠ACB intersects the circle at point D. Prove that, seg AD ≅ seg BD. Complete the following proof by filling in the blanks.

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