In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a : Options: A . &nbspRight isosceles triangle B . &nbspIsosceles triangle C . &nbspRight-angled triangle D . &nbspEquilateral triangle

In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a :

Question

In ΔABC, AB = BC = K, AC = $$\sqrt 2 $$ k, then ΔABC is a :

Options:

A . Right isosceles triangle

B . Isosceles triangle

C . Right-angled triangle

D . Equilateral triangle

Answer: Option A

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