Question

In the given 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.

Answer 1

Proof: Draw seg OD.

∠ACB = $\boxed{90°}$ ..... angle inscribed in semicircle

∠ACB = $\boxed{45°}$ ..... CD is the bisector of ∠C

m(arc DB) = 2∠ACB = $\boxed{90°}$ .....inscribed angle theorem

∠DOB = $\boxed{90°}$ .....definition of measure of an arc (I)

seg OA ≅ seg OB .....$\boxed{\mathrm{Radii}\mathrm{of}\mathrm{the}\mathrm{circle}}$ (II)

∴ line OD is $\boxed{\mathrm{perpendicular}\mathrm{bisector}}$ of seg AB .....From (I) and (II)

∴ seg AD ≅ seg BD