Final answer:
Triangles ADE and ABCF in an isosceles trapezium are similar by the Angle-Angle criterion since they share two pairs of equal angles. The correct choice that supports this similarity is option C: AADE ∼ ABCF.
Step-by-step explanation:
The problem presents us with an isosceles trapezium ABCD, where AB and DC are parallel sides, and perpendiculars DE and CF are drawn to AB. To check whether triangles ADE and ABCF are similar, we would use triangle similarity criteria such as AA (Angle-Angle). In an isosceles trapezium, the angles adjacent to each base are equal, so we have ∠ADC = ∠BCA and ∠CAB = ∠DAB. Since DE and CF are both perpendicular to AB, they create right angles; therefore ∠ADE = ∠ACF = 90 degrees. Hence, we have two angles equal in each triangle which shows that △ADE is similar to △ACF by AA criterion.
However, the letter options provided in the question (A. DC ∼ CA B. DE ∼ CF C. AADE ∼ ABCF D. ABCD ∼ DCBA) do not directly correspond to the criteria for triangle similarity or congruence. Nonetheless, statement C suggests the similarity between triangles ADE and ABCF, which can be supported by the AA similarity noted above, so option C is the correct one.
Option A is incorrect because DC and CA are not triangles and thus cannot be similar. Option B mistakenly equates line segments DE and CF, which are not necessarily equal in length even if the triangles they belong to are similar. Lastly, option D mistakenly suggests that the whole trapezoid ABCD is similar to itself flipped, which is true but does not relate to the question of the similarity between triangles ADE and ABCF.