#### Question

AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.

#### Solution

Join OC,

`∠`BCD = `∠`BAC = 30° (angles in alternate segment)

Arc BC subtends `∠`DOC at the centre of the circle and `∠`BAC at the remaining part of the circle.

∴ `∠` BOC = 2`∠`BAC = 2 × 30° = 60°

Now in Δ OCD,

`∠`BOC or `∠`DOC = 60°

`∠`OCD = 90° (OC ⊥ CD)

∴ `∠`DCO+ `∠`ODC = 90°

⇒ 60° + `∠`ODC = 90°

⇒ `∠`ODC = 90° - 60° = 30°

Now in ΔBCD,

∵ `∠`ODC or `∠`BDC = `∠`BCD = 30°

∴ BC = BD

Is there an error in this question or solution?

Solution Ab is the Diameter and Ac is a Chord of a Circle with Centre O Such that Angle Bac = 30°. the Tangent to the Circle at C Intersects Ab Produced in D. Show that Bc = Bd. Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.