Find AB and BC, if :

Let BC be x cm.

BD = BC + CD = (x + 20) cm

In Δ ABD,

$\text{tan 30°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AB}{BD}\\[1em] ⇒ \dfrac{1}{\sqrt3} = \dfrac{AB}{x + 20}\\[1em] ⇒ x + 20 = AB\sqrt3 …………..(1)$

In Δ ABC,

$\text{tan 45°} = \dfrac{Perpendicular}{Base}\\[1em] ⇒ 1 = \dfrac{AB}{BC}\\[1em] ⇒ 1 = \dfrac{AB}{x}\\[1em] ⇒ x = AB …………..(2)$

From equation (1),

AB + 20 = AB $\sqrt3$

⇒ AB $\sqrt3$ - AB = 20

⇒ AB($\sqrt3$ - 1) = 20

⇒ AB(1.73 - 1) = 20

⇒ AB x (0.73) = 20

⇒ AB = $\dfrac{20}{0.73}$

⇒ AB = 27.32 cm

And BC = x = AB = 27.32 cm

Hence, AB = BC = 27.32 cm.