Question

In the adjoining 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 the blanks.

Answer 1

Given: seg AB is a diameter, seg CD bisects ∠ACB.

To prove: seg AD ≅ seg BD 

Construction: Draw seg OD. 

Proof: 

∠ACB = 90° [Angle inscribed in a semicircle] 

∠DCB = ∠DCA = 45° [CD is the bisector of ∠C] 

m(arcDB) = 2∠DCA = 90° [Inscribed angle theorem] 

∠DOB = m(arc DB) = 90° ………… (i) [Definition of measure of arc]

segOA ≅ segOB …………. (ii) [[Radii of the same circle] 

∴ line OD is the perpendicular biscctor of [From (i) and (ii)] 

seg AB.

∴ seg AD ≅ seg BD