Question

AB is diameter of a circle
with centre 0. The chord
BC of the circle is parallel
to the radius. OD. Prove
that
() angle CED = 3 angle CBD
(II) CD = DA​

Answer 1

Step-by-step explanation:

(a)

Arc CD subtends ∠COD at the centre and ∠CBD at remaining part of circle.

∴ ∠COD = 2∠CBD      ------ (1)

∵ BC ║ OD

∴ ∠CBD = ∠BDO     ----- (2)

In ΔDOE,

∠BEO = ∠EDO + ∠EOD

          = ∠BDO + ∠COD

          = ∠CBD + 2 ∠CBD

          = 3∠CBD

But,

∠CED = ∠BEO(Vertically opposite angles)

∴ ∠CED = 3∠CBD

(b)

In ΔDBO,

OD = OB (radii of same circle)

∴ ∠OBD = ∠BDO

             = ∠CBD  {From (2)}

⇒ ∠ABD = ∠CBD

{∵Equal chord subtend equal angles}

∴ AD = CD

Hope it helps!