The length of BK = a√3/2 and the area of △ABC = a²√3 / 4.
Since AB = BC = AC = a, △ABC is an equilateral triangle.
In an equilateral triangle, the altitude drawn from any vertex to the opposite side bisects that side and intersects the base at the midpoint.
BK bisects AC, dividing it into two segments of length a/2.
In an equilateral triangle, the height drawn from any vertex to the opposite side also bisects the base angle, creating two congruent 30-60-90 triangles.
In this case, BK bisects ∠C into two 30° angles.
In a 30-60-90 triangle, the shorter leg is half the hypotenuse and the longer leg is √3 times the shorter leg.
Therefore, in △ABK, AK = a/2 and BK = (√3)(a/2) = a√3/2.
The area of a triangle is calculated as (base x height)/2.
In △ABC, the base is AC = a, and the height is BK = a√3/2.
Therefore, the area of △ABC is:
Area = (a x a√3/2) / 2 = a²√3 / 4