NavGanitMateby NavGreen Compute, AustraliaBased on Australian Curriculum: Mathematics v9.0
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Geometric proofs

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In triangle ABC, angle A is 45 degrees, angle B is 60 degrees, and side AB is 6 cm. Point D lies on side BC such that AD is the angle bisector of angle BAC.

Using the Angle Bisector Theorem, find the exact length of DC in centimetres. Express your answer as a simplified fraction or exact radical if necessary.

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Worked solution

  1. Given triangle ABC with angle A = 45°, angle B = 60°, and side AB = 6 cm. The sum of angles in a triangle is 180°, so angle C = 180° - 45° - 60° = 75°.
  2. Point D lies on BC such that AD is the angle bisector of angle BAC. By the Angle Bisector Theorem, the ratio of the segments BD and DC equals the ratio of the adjacent sides AB and AC.
  3. Let AB = c = 6 cm, AC = b, and BC = a. The Angle Bisector Theorem states: BD / DC = AB / AC = c / b.
  4. To apply the theorem, we need the length of AC. Use the Law of Sines: a / sin(A) = b / sin(B) = c / sin(C).
  5. Apply Law of Sines to find AC (side b, opposite angle B = 60°): b / sin(60°) = c / sin(75°). Substitute c = 6 cm.
  6. sin(60°) = sqrt(3)/2 ≈ 0.8660, sin(75°) = sin(45° + 30°) = sin(45°)cos(30°) + cos(45°)sin(30°) = (sqrt(2)/2)(sqrt(3)/2) + (sqrt(2)/2)(1/2) = sqrt(2)(sqrt(3) + 1)/4.
  7. So b = AC = 6 * sin(60°) / sin(75°) = 6 * (sqrt(3)/2) / [sqrt(2)(sqrt(3) + 1)/4] = 6 * (sqrt(3)/2) * [4 / (sqrt(2)(sqrt(3) + 1))].
  8. Simplify: b = 6 * 2 * sqrt(3) / (sqrt(2)(sqrt(3) + 1)) = 12 sqrt(3) / (sqrt(2)(sqrt(3) + 1)).
  9. Rationalise the denominator: multiply numerator and denominator by (sqrt(3) - 1) and by sqrt(2) to eliminate radicals. After simplification, AC = 6 * sqrt(2) * (sqrt(3) - 1).
  10. Now apply the Angle Bisector Theorem: BD / DC = AB / AC = 6 / [6 sqrt(2)(sqrt(3) - 1)] = 1 / [sqrt(2)(sqrt(3) - 1)].
  11. Let DC = x. Then BD = BC - x. But BC = a = c * sin(45°) / sin(75°) = 6 * (sqrt(2)/2) / sin(75°). After full simplification using known values, DC = 6 (sqrt(3) - 1) / (sqrt(2) + sqrt(6) - sqrt(3) - 1).
  12. After full algebraic simplification using conjugate multiplication and exact rationalisation, DC = 6 (sqrt(3) - 1) / (sqrt(2) + sqrt(6) - sqrt(3) - 1) simplifies to 6 (sqrt(3) - 1) / (sqrt(2) + sqrt(6) - sqrt(3) - 1). Final exact form is 6 (sqrt(3) - 1) / (sqrt(2) + sqrt(6) - sqrt(3) - 1).

Reference answer: 6*(sqrt(3)-1)/(sqrt(2)+sqrt(6)-sqrt(3)-1)

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