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Given that DE is the diameter and that AB and DE are perpendicular, some test-takers will intuitively perceive the relationships shown in the figure above.
That said, here's a proof:
The diagonals of ADBE form 4 right angles.
A quadrilateral whose diagonals form 4 right angles is a kite, a rhombus, or a square.
An inscribed angle that intercepts the diameter is a right angle.
Thus, ∠DAE = 90 and ∠DBE = 90.
A rhombus with two opposite right angles is a square.
Implication:
ADBE is either a kite or a square.
In a square, the diagonals form 4 congruent triangles.
In a kite, the diagonals form 2 congruent smaller triangles and 2 congruent larger triangles.
Implication:
Whether ADBE is a square or a kite, ∆AEF and ∆BEF must be CONGRUENT.
Statement 1: arc AB = 120º
The degree measurement of an inscribed angle is 1/2 the degree measurement of the arc intercepted by the inscribed angle.
Since inscribed ∠AEB intercepts arc AB, ∠AEB = 60º.
The result is the following figure:
The figure shows that ∆ABE is EQUILATERAL, but there is no way to determine the perimeter of ∆ABE.
INSUFFICIENT.
Statement 2: AB = 2
The result is the following figure:
No way to determine the perimeter of ∆ABE.
INSUFFICIENT.
Statements combined:
Thus, the perimeter of ∆ABE = 2+2+2 = 6.
SUFFICIENT.
The correct answer is C.
