Given:

OA = 5 cm

OQ = 3 cm

Property 1: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 2: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

By property 1, ∆OAQ is right-angled at ∠OQA (i.e., ∠OQA = 90°).

By Pythagoras theorem in ∆OAQ,

OA² = QA² + OQ²

⇒ QA² = OA² – OQ²

⇒ QA² = 5² – 3²

⇒ QA² = 25² – 9²

⇒ QA² = 16

⇒ QA = √16

⇒ QA = 4 cm

By property 2,

BQ = BP (tangent from B)

And,

AQ = BQ = 4 cm [∵ Q is midpoint of AB]

PB = PC = 4 cm [∵ P is midpoint of BC]

Now,

BC = BP + PC

⇒ BC = 4 cm + 4 cm

⇒ BC = 8 cm

Hence, BC = 8 cm