In the diagram, two circles intersect at #B# and #E#. #ABC# is a line parallel to the #FED#. How do you prove that #ACDF# is a parallelogram?

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1 Answer
Nov 19, 2017

Proof: involving co-interior angles and cyclic quadrilaterals

Explanation:

Since we already know that #ABC# is parallel to #FED#, all we need to do is to prove that the opposite angles are equal.
We're going to do this by finding expressions for the angles #EhatFA, FhatAB, BhatCD# and #ChatDE#.

Since ABC and FED, are parallel, the angles form between those and FA and CD are co-interior (also know as allied) angles.

Two co-interior angles sum to 180.
#:. EhatFA+FhatAB=180#

What we have here with ABEF and BCDE are cyclic quadrilaterals; since all of the sides of the quadrilaterals they form lie on the circumference of the same circle.

Opposite angles in a cyclic quadrilateral sum to 180
#:. EhatFA+AhatBE=180#

We now have two simultaneous equations involving #EhatFA#, which we can solve. I solve these by eliminating #EhatFA#, but you can also set these equal to each other since they both equal 180.

#EhatFA+FhatAB=180-#
#EhatFA+AhatBE=180#

#FhatAB-AhatBE=0#
#FhatAB=AhatBE#

You should get this no matter which one you use.

Let #FhatAB=AhatBE=x#
(this will make calculations from now on easier to write).

#EhatBC=180-x# from angles on a straight line
#ChatDE=x# opp. angles in a cyclic quad. (opposite to EBC).
#FhatAB=x# as we defined before.

#FhatAB=ChatDE# which are opposite each other.

#BhatCD=180-x# since co-interior to #ChatDE#
#EhatFA=180-x# since co-interior to #FhatAB#

#:.EhatFA=BhatCD# which are opposite each other.

#:.# opposite angles in the quadrilateral #ABCDEF# are equal
#:.ABCDEF# is a parallelogram

#Q.E.D.#