mathematical help please?
Parallelogram ACFD is split in two parts so that ABED and FEBC are congruent isosceles trapezoids. What are the measures of all the angles of trapezoid FEBC if angle D is 48°?
please answer this for me because i am clueless on how to solve it.
- bpiguyLv 78 years agoFavorite Answer
The existence of parallelogram ACFD almost doesn't matter. The main thing is that ABED and FEBC are congruent isosceles trapezoids. (Notice that they share a common side BE that's parallel to and equidistant from AD and CF.)
Anyway, ∠D = ∠A = 48° (given; base ∠s of isos. trap. are =). ∠BED = ∠EBA = 180 - 48 = 132° (base ∠s =; 360° in quadrilateral).
Applying these measures to corresponding angles in the congruent trapezoid, we have ∠C = ∠F = 48° (Answer) and ∠FEB = ∠CBE = 132° (Answer).