Question

In the figure below AOD is a diameter of the circle cetre O. BC is a chord parallel to AD. FE is a tangent to the circle. OF bisects angle COD. Angle BCE = angle COE = 20° BC cuts OE at X

Calculate;
(a) angle BOE
(b) angle BEC
(c) angle CEF
(d) angle OXC
(e) angle OFE​

Answer 1

Explanation:

∠ACB=90∘

[∠ from diameter]

In ΔACB

∠A+∠ACB+∠CBA=180∘

∠CBA=180∘

−(90+30)

∠CBA=60∘ (1)

In △OCB

OC=OB

So, ∠OCB=∠OBC

[The sides are equal]

∠OCB=60∘

∠OCD=90∘

∠OCB+∠BCD=90°

∠BCD=30∘ (2)

∠CBO=∠BCO+∠CDB

[external ∠ bisectors]

60=30+∠CDB

∠CDB=30° (3)

from (2) & (3)

BC=BD

[The ∠.S are equal]