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.
Statement | Reason |
---|---|
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.