The diagram shows a circle, centre OO, with diameter ABAB. The points DD and FF lie on the circle. The point EE is such that EBEB and EFEF are tangents to the circle.

  1. (a)

    Given that the points CC and DD are midpoints of BEBE and AEAE respectively, prove that angle DCE=90∘DCE = 90^\circ.

  2. (b)

    Given that triangle BEFBEF is equilateral, prove that ∠BEF=∠BAF\angle BEF = \angle BAF.

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Worked solution

3(a)

1. Identify the angle between the diameter and the tangent.

Since ABAB is a diameter and EBEB is a tangent to the circle at BB,
∠ABE=90∘\angle ABE = 90^\circ (tangent ⊥\perp radius).

2. Apply the Midpoint Theorem to triangle ABE.

In △ABE\triangle ABE, DD is the midpoint of AEAE and CC is the midpoint of BEBE.
By the Midpoint Theorem, CD∥ABCD \parallel AB.

3. Use parallel lines to find angle DCE.

Since CD∥ABCD \parallel AB and EBEB is a transversal,
∠DCE=∠ABE\angle DCE = \angle ABE (corresponding angles).
Therefore, ∠DCE=90∘\angle DCE = 90^\circ.

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3(b)

1. Identify the angles in the equilateral triangle

Since triangle BEFBEF is equilateral, ∠EFB=∠BEF=60∘\angle EFB = \angle BEF = 60^\circ.

2. Apply the alternate segment theorem to relate the angles

Since EFEF is a tangent to the circle at FF and FBFB is a chord, by the alternate segment theorem, the angle between the tangent and the chord is equal to the angle in the alternate segment.
Therefore, ∠BAF=∠EFB\angle BAF = \angle EFB.
Since ∠EFB=∠BEF\angle EFB = \angle BEF, it follows that ∠BAF=∠BEF\angle BAF = \angle BEF.

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