In $$\triangle ABD, C$$ is the midpoint of $$BD$$. If $$AB = 10 cm, AD = 12 cm$$ and $$AC = 9 cm$$, then $$BD = ?$$

Let the BC = CD = x cm.
BD = 2x

According to heron's formula, the area of Δ ABD is:

s = $$\frac{a+b+c}{2}$$

s =  $$\frac{10+12+2x}{2}$$ = 11 + x

a = 10 cm, b = 12cm, c = 2x cm

area = $$\sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{(11 + x)(11 + x-10)(11 + x-12)(11 + x-2x)} = \sqrt{(11 + x)(1 + x)(x - 1)(11 - x)}$$

= $$\sqrt{(121 - x^2)(1 - x^2)}$$
Similarly in $$\triangle ABC$$
s = $$\frac{10+9+x}{2} = \frac{19 + x}{2}$$
Area of \triangle ABC = \sqrt{s(s-a)(s-b)(s-c)}$$
= $$\sqrt{(\frac{19 + x}{2})(\frac{19 + x}{2}-10)(\frac{19 + x}{2}-9)(\frac{19 + x}{2}-x)}$$

=$$\sqrt{\frac{(361 - x^2)(x^2 - 1)}{16}}$$
AC is a median so,
Area of $$\triangle ABC =(1/2) Area of \triangle ABD$$

$$\sqrt{\frac{(361 - x^2)(x^2 - 1)}{16}}$$ = (1/2) $$\times \sqrt{(121 - x^2)(1 - x^2)}$$

$$\frac{(361 - x^2)(x^2 - 1)}{16}$$ = (1/4) $$\times {(121 - x^2)(1 - x^2)}$$

$$(361 - x^2)(x^2 - 1) = 4 \times (121 - x^2)(1 - x^2)$$

$$361 - x^2 = 484 -4x^2$$

$$x^2 = 41$$

x = $$\sqrt{41}$$

BD = 2x = 2$$\sqrt{41}$$