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.
Proof: Draw seg OD.
…90° ............ angle inscribed in semicircle
........45° .... CD is the bisector of 
m(arc DB) = 
........45° .. inscribed angle theorem
…............90° definition of measure of an arc (I)
seg OA ≅seg OB .......radii of the circle... 
(II)
∴line OD is 
of seg AB ......bisector.... From (I) and (II)
∴seg AD ≅seg BD
Proof: Draw seg OD.
∠ ACB = 90° {angle inscribed in semicircle}
∠ DCB = 45° {CD is the bisector of 
}
m(arc DB) = 45° {inscribed angle theorem}
∠ DOB = 90° {definition of measure of an arc} (I)
seg OA ≅seg OB {radii of the circle}(II)
∴line OD is bisector of seg AB From (I) and (II)
∴seg AD ≅seg BD