A triangle ABC is inscribed in a circle. A tangent is drawn to the circle at point A, intersecting the extension of side BC at D. If AD=6 cm and CD=4 cm, what is the length of side BC?

Correct — A. D lies on BC extended beyond C, so the line from D meets the circle first at C, then at B.

Tangent–secant rule: DA² = DC × DB
Square the tangent: 6² = 36
Divide by DC: DB = 36 ÷ 4 = 9 cm

Whole secant: DB = DC + CB
Subtract: BC = 9 − 4 = 5 cm → option (a)

Why the others are wrong

• (b)
  6 cm — BC = 6 cm makes DB = 4 + 6 = 10, so DC × DB = 40, not the 36 that a 6 cm tangent requires. 6 cm is the tangent AD, not the chord.
• (c)
  7 cm — BC = 7 cm makes DB = 11, so DC × DB = 44 and the tangent would be √44 ≈ 6.63 cm, not 6 cm.
• (d)
  8 cm — BC = 8 cm makes DB = 12, so DC × DB = 48 and the tangent would be √48 ≈ 6.93 cm. A 6 cm tangent needs DB = 9 cm.

Tangent–secant theorem: from a point outside a circle, the square of the tangent equals the outside part of a secant times the whole secant. Here DA² = DC × DB.

Both lengths on the right are measured from D. DC stops at the near point, DB runs on to the far point, and the chord BC is the difference, DB − DC.

Key facts

• Tangent–secant theorem: (tangent)² = outside part × whole secant, both measured from the outside point.
• The chord cut off inside the circle = whole secant − outside part.
• Every secant through the same outside point gives the same product, equal to the tangent squared.

Common traps

• Multiplying the outside part by the chord, DC × CB = 36, instead of by the whole secant DB. That gives 9 cm, which is not offered.
• Stopping at DB = 9 cm, the whole secant, without subtracting DC = 4 cm.

No directly related past PYQ was found.