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

Solution (By Examveda Team)

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
$$\eqalign{ & \cos \theta = \frac{{{\text{B}}{{\text{C}}^2} + {\text{C}}{{\text{D}}^2} - {\text{B}}{{\text{D}}^2}}}{{2{\text{BC}}.{\text{CD}}}} \cr & \frac{4}{5} = \frac{{{{80}^2} + {{40}^2} - {{\text{y}}^2}}}{{2 \times 40 \times 80}} \cr & \frac{4}{5} \times 80 \times 80 = 6400 + 1600 - {{\text{y}}^2} \cr & 5120 = 8000 - {{\text{y}}^2} \cr & {{\text{y}}^2} = 2880 \cr & {\text{y}} = 24\sqrt 5 {\text{ cm}} \cr} $$