In ΔABC, Perimeter of ΔABD = Perimeter of ΔBCD and AB = 60 cm, BC = 80 cm and AC = 100 cm. Then find BD.

Answer: Option C

Let AD = x and BD = y
Perimeter of ΔABD = Perimeter of ΔBCD
60 + x + y = 80 + y + 100 - x
2x = 120
x = 60
CD = 100 - 60 = 40
Let ∠ACB = θ
In ΔABC,
cosθ $$=\frac{80}{100}=\frac{4}{5}$$
In ΔCBD, by using cosine formula we get
$$\begin{array}{rl}& \cos \theta =\frac{\text{B}{\text{C}}^2+\text{C}{\text{D}}^2-\text{B}{\text{D}}^2}{2\text{BC}.\text{CD}}\\ & \frac{4}{5}=\frac{{80}^2+{40}^2-{\text{y}}^2}{2\times 40\times 80}\\ & \frac{4}{5}\times 80\times 80=6400+1600-{\text{y}}^2\\ & 5120=8000-{\text{y}}^2\\ & {\text{y}}^2=2880\\ & \text{y}=24\sqrt{5}\text{ cm}\end{array}$$