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

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 -

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Try showing #3 first. – tskuzzy Jun 19 '12 at 21:32

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

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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.

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So from Thales it states that AC/CE = BD/DE . How does this help when trying to show that AB is double CD ? – MistyD Jun 19 '12 at 22:05
You only need the cosine rule to show that AB is double CD. – Peter Phipps Jun 19 '12 at 23:09
I need it done without Trigonometric ratio sorry i didn't mention that in the question. – MistyD Jun 20 '12 at 0:42
Oh, if you need to solve it only with trigonometry, then Peter is right: use the cosine rule (aka Carnot's theorem) to prove that AB is twice CD (point 1). Then you have that all the sides of ABE are twice the sides of CDE (point 3). Using again cosine rule (or even sine rule) you can show that the angles ECD and EAB are equal, and same for EDC and EBA (point 2). – bartgol Jun 20 '12 at 13:57