Question

In the figure, 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.

∠ACB = `square`   ...[Angle inscribed in semicircle]

∠DCB = `square`   ...[CD is the bisector of ∠C]

m(arc DB) = `square`   ...[Inscribed angle theorem]

∠DOB = `square`   ...[Definition of measure of an arc] (i)

seg OA ≅ seg OB   ...`square` (ii)

∴ Line OD is `square` of seg AB   ...[From (i) and (ii)]

∴ seg AD ≅ seg BD

Answer 1

∠ACB = \[\boxed{90°}\]   ...[Angle inscribed in semicircle]

∠DCB = \[\boxed{\text{∠DCA} = 45°}\]   ...[CD is the bisector of ∠C]

m(arc DB) = \[\boxed{\text{2∠DCB} = 90°}\]   ...[Inscribed angle theorem]

∠DOB = \[\boxed{\text{m(arc DB)} = 90°}\]   ...[Definition of measure of arc] (i)

seg OA ≅ seg OB   ...\[\boxed{\text{[Radii of the same circle]}}\] (ii)

∴ Line OD is \[\boxed{\text{perpendicular bisector}}\] of seg AB   ...[From (i) and (ii)]

∴ seg AD ≅ seg BD