Need a little help understanding this triangle concept. - Thales Theorem

Question

The question is:

In the following diagram AC=CE and BD=DE .Which of these statements are true ?
1-AB is twice as long as CD
2-AB is parallel to CD
3-Triangle AEB is similar to triangle triangle CED
Note: All these statements are correct.

Could anyone explain how the three statements are correct without using Without Trigonometric ratios especially if there is no link between AE and BE

Edit: I assume we need to use Thales theorem here -

Answer 1

All of these follow from Thale's Theorem (an old and well known result in geometry), that you can find here.

In particular, statement $(3)$ follows immediately from the assumptions (and one of the similarity criteria for triangles). Then you can prove that CD is parallel to AB and invoke Thale's theorem to prove the other two.

Answer 2

From Bartgol suggestions we could use Thales Theorem.

For Point 1:

According to Thales theorem: $\dfrac{AB}{CD} = \dfrac {BE}{DE}$

$\dfrac{AB}{CD} = 2\cdot \dfrac{DE}{DE}$

$AB=2\cdot CD$

For Point 2: Again using Thales theorem $\dfrac {AC}{AE} = \dfrac{BD}{BE}$ (Hence Parallel) and Similar