Question

In the figure ,seg AB is the diameter of the circle with center O. The bisector of ∟ACB intersects the circle at pt D. Prove that , seg AD ≅ seg BD​

Answer 1

Answer:

Draw seg OD.

∠ACB=90 °

....angle inscribed in semicircle

∠ACB=45 °

CD is the bisector of ∠C

m(arc DB)=2∠ACB=90 °

inscribed angle theorem

∠DOB=90 °

.... definition of measure of an arc (I)

seg OA ≅ seg OB Radii of the circle (II)

∴ line OD is perpendicular bisector of seg AB From (I) and (II)

∴ seg AD ≅ seg BD