Since l is a bisector, point E is in the middle of AB. This tells us that at least one point on l is in the same distance from point A as from point B. Now, let's choose a point X on line l, which is different than E.
Now, let's analyze â–³ AEX and â–³ EBX.
Since E is the midpoint of AB, we get that AE=EB. Furthermore, EX is a common side of △ AEX and △ EBX. Notice that the angles ∠AEX and ∠BEX are right. Therefore, by the Side-Angle-Side Similarity Theorem triangles △ AEX and △ EBX are congruent.
Since â–³ AEX and â–³ EBX are congruent, AX=BX. Therefore, point X is in the same distance from point A as from point B. This tells us that all points on l are the same distance from point A as from point B. This corresponds to option E.
