In an irregular trapezium of ABCD with diagonals that intersect at point E in the middle, are all the triangles similar?

Question

In this diagram above, let's imagine this was an irregular trapezium. This means that none of the four sides are the same length.

I know that AB is parallel to DC.

<dec = bea = vertically opposite

<abe = cde = alternate angles

Therefore by AA we can show that these triangles are similar.

Without measuring or anything is it possible for <ade and <bce to be similar?

My understanding is that they would be similar because if AD = BC then it would be congruent. But because it is an irregular trapezium of each side having varying lengths then it would be similar because the angle should be the same?

Answer 1

Since $\angle AED = \angle BEC$, the two triangles $AED$ and $BEC$ are similar if and only if:

• either $\angle DAE = \angle BCE$,
• or $\angle DAE = \angle CBE$.

In the first case, $AD$ is parallel to $BC$ and we get a parallelogram.

In the second case, we get a cyclic quadrilateral, and it is easy to see that a cyclic trapezium is necessarily isosceles.

In both cases, we would have $|AD| = |BC|$, which is excluded by assumption.