What if I showed angles DEB and EBC to be equall

[m00c0w]

http://img298.echo.cx/img298/2145/pentagon8zw.png ABCDE is a regular pentagon, in the previous question I worked out angle AEB to be 36°.. the next question has asked to prove that BCDE is a trapezium but I don't know how to. I figured it might be to do with showing that the sum of the angles in BCDE = 360° but that doesn't necessarily prove it is a trapezium. What if I showed angles DEB and EBC to be equal.. does that prove it?

Many thanks.

 

[BicycleTree]

You want to show that lines EB and DC are parallel. One way: pick a point F to the left of D along line CD, and show DEB = EDF.

 

[steven187]

hello there

a trapezium needs to have two sides parallel to each other and the other two sides unparallel to each other, if you want you could draw diagnals through the trapezium and prove that the alternate angles are equal that is prove that angle EBD= BDC
or anything similar to prove that it has two parallel sides

check out this simple link
http://www.icteachers.co.uk/children/sats/quadrilaterals.htm

Steven

 

[BicycleTree]

I have never heard that a trapezoid (trapezium) had to have the other two sides non-parallel. Mathworld and dictionary.com are ambiguous but wikipedia is clear that a trapezoid can also be a parallelogram.
http://en.wikipedia.org/wiki/Trapezoid

 

[steven187]

I have never heard that a trapezoid (trapezium) had to have the other two sides non-parallel. Mathworld and dictionary.com are ambiguous but wikipedia is clear that a trapezoid can also be a parallelogram.
http://en.wikipedia.org/wiki/Trapezoid

I knew i should have not erased what i originally wrote, well anyway i provided the link so that he doesn't get confused, well such a shape has a proper name a parallelogram which is a part of the family of quadrilaterals, and in this case all that is necessary is to prove that it has one pair of parallel sides to show that it is a trapezium