garaidh wrote:
Certainly. Since AB=DE, CD || BE, and BC || DF, we have BC=CF=DF=CD due to the statement "If two parallel lines intersect two parallel lines, each of the resulting segments will be equal to their opposite." This is equivalent to the Euclidean parallel postulate (which we are allowed to assume on all GRE questions). It should not be two hard to prove this or find a proof of this if you are unconvinced, but it should make sense intuitively because if it were the case that BF > DF, we would necessarily have BF (BE) intersecting CD, which is impossible by assumption CD || BE. Thus it must be the case that BC = DF and BF = CD and since we know BC = CD, we have by transitivity of equality that BF = DF (that is, they are all equal to each other).
As for your second question, we do know that BCDF is a parallelogram because CD || BF and BC || DF. Furthermore, since we know by the above reasoning that CD=DF, if we were to draw a circle of radius CD centered at D, it would intersect both C and F. We can then draw another circle of the same radius centered at B, which will also intersect C and F. You might recall that this is the construction of a perpendicular to BD. Now, we know that AB || DE and that AB=DE. We then have that BD || AE (again by statement equivalent to Euclid's Parallel Postulate), thus CF must be perpendicular with AE. We have CF || DE because both make right angles with AE and both intersect the parallel lines BE and CD. Then by Euclid's Fifth, we have CF=DE.
HI,
As per Euclidean parallel postulate, the distance between two parallel lines at point will always be equal, so yes we can concur that CD =BF and BC = FD, but since BC = CD, we can write as BC = CD = DF = FB,
This led us that BCDF is a rhombus, moreover the diagonal bisect each perpendicularly so CF and BD will be perpendicular and if we consider two triangles BCD and BDF, they will always have equal area.
The only confusion with the statement
CF || DE,
However I approached a different way,
SInce AB || DE and BD || AE also AB= DE and BD = AE (Euclidean parallel postulate), so ABDE is a parallelogram and the diagonals of the parallelogram divides it into 4 triangles of equal area.
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