In this problem, show that triangles BAE, CDE and FEC are all similar.
One method to show this is to prove that corresponding angles of each triangle are equal.
|Triangles BAE, CDE and FEC|
|DC || AB||Since DC and AB are extended (base)lines of a parallel trapezoids.|
|Angle A and angle D are both right angles||Intersection of altitude and a horizontal line forms a right angle.|
|Angle C and angle B are congruent||Parallel lines cut by a tranversal lines - corresponding angles.|
|Thus triangle BAE and CDE are similar||AA or AAA theorem|
|Angle E is a right angle||Line CE is the distance between two parallel lines.|
|∠ F + 90 + \(\phi\) = 180||Sum of interior angles of a triangle is 180|
|∠ C + 90 + \(\phi\) = 180||Supplementary angles|
|∠ F = ∠ C||Simplifying the previous two equations|
|∠ D = ∠ = E = 90||All 90 are equal|
|Triangle FEC and CDE are similar||AA theorem|
|The three triangles are similar||Transitivity|
** Use the given figure for clarity and other missing information.