In ΔABC and ΔA’B’C’

Also, In ΔABC

∠A + ∠B + ∠C = 180°

⇒ ∠A + 30° + 20° = 180°

⇒ ∠A = 130°

So here,

∠A = ∠A’

⇒ ΔABC ∼ ΔA’B’C’ [By SAS Similarity]

⇒ t = 8 units

OR

By Pythagoras theorem,

(Hypotenuse)² = (Base)² + (Perpendicular)²

AB² = AC² + BC²

[Given: AC = BC, as it is an isosceles triangle]

AB² = 2AC²

Hence, Proved.