Question

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?

Answer 1

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.`