In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.

Join OE and AB.

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

We know that,

Angle at the centre is twice the angle at remaining circumference.

∴ ∠EOC = 2∠EBC = 2 x 65° = 130°.

Now, in ∆OEC

OE = OC [Radii of the same circle]

So, ∠OEC = ∠OCE [Angle opposite to equal sides are equal.]

In ∆OCE by angle sum property,

⇒ ∠OEC + ∠OCE + ∠EOC = 180°

⇒ 2∠OCE + 130° = 180°

⇒ 2∠OCE = 180° - 130°

⇒ 2∠OCE = 50°

⇒ ∠OCE = $\dfrac{50°}{2}$ = 25°.

Given, AC || ED

∴ ∠DEC = ∠OCE [Alternate angles are equal]

⇒ ∠DEC = 25°.

Hence, ∠DEC = 25°.