(a)  Given △ABC is similar △BDC

Then    AB/BD = BC/DC = AC/BC

DC = CD = 6,   AD = 18   => AC =18+6 = 24

BC/6 = 24/BC

(BC)² =24 x 6 = 144   => BC = CB = 12cm

(b)  Let Angle BAC = ∠A,  a = BC = 12 , b = AC =24, c = AB = 15

Using Cosine Rule ,  a = √[c² + b² – 2ab cos A] =>  a² = [c² + b² – 2ab cos A]

– 2bc cos A = a²  –  c² – b²

-(2)(24)(15) cos A = 144 – 225 – 576 = -657

cos A = 0.9125

∠A = Angle BAC = 24.15⁰ (to 4 s.f.)

(c) Using the Sine Rule , Area of  △ABC = (1/2) bc sin A = (1/2)(24)(15) sin 24.15 = 73.64

Area of △BDC / Area of △ABC =  (BC/AC)²  

Area of △BDC = (12/24)² x 73.64 = 18.4 cm²  (to 3 s.f.)

Thank you very much.

sushi88 already gave an excellent answer. It is appropriate for Sec 3, where students should be aware of Cosine Rule and use it accordingly.

Below, for fun, I want to pretend I am only Sec 1 or 2 and only know definition of cosine and Pythagoras Theorem.

Solution using Pythagoras Theorem and basic cosine

Part (b):
After part (a), we know triangle ABC has sides 15, 12 and 24.
Let h be the perpendicular. Then h2 is
     152 - x2 = 122 - (24-x)2
    152 - 122 = x2 - (24-x)2
       27 * 3 = 24 * (2x - 24)   (difference of 2 squares on both sides!)
            x = (81/24 + 24)/2 = 13.6875
So angle BAC = cos-1 (x/15) = 24.15 degrees (2 decimal places).

Part (c): 
We don't even need angle BAC or knowledge of cosine. Just 1/2 base * height.
  Area of ABC = 12 * h = 12 * sqrt(152 - x2)
              = 73.6338 (4 decimal places)
  Area of BCD = (6/12)**2 * Area ABC = 18.41 (2 decimal place)

Part (c) alternative:
In fact, if there was no Part (b), we can solve (c) using Heron's Formula.
The sides are a = 12, b = 15, c = 24. So s = (a+b+c)/2 = 25.5.
Area ABC = sqrt(s(s-a)(s-b)(s-c)) = 73.6338 (4 dp), etc.
Heron's Formula is probably not in the syllabus but it is useful to know. :)