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.