Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm looking at this link:

..and it says "Called an Isosceles trapezoid when the sides that aren't parallel are equal in length and both angles coming from a parallel side are equal."

Isn't the second part, namely "both angles coming from a parallel side are equal" redundant? I mean, if the two sides that aren't parallel are equal in length, then doesn't it follow that the angles will be equal as well? I can't prove this, but my intuition tells me this is how it is. Am I correct in making this assumption?

share|cite|improve this question

Yes, you are correct. Given any trapezoid, we can determine whether or not it is also isosceles by using any number of necessary and sufficient characterizations, including "equal legs" and "equal base angles".

To see why "equal legs" implies "equal base angles", imagine extending the legs (that is, extending the sides that aren't parallel) until they form a triangle. Notice that, because trapezoids have one pair of parallel sides, there are now two similar triangles. Since the trapezoid has two equal legs, the larger triangle must also have two equal legs. Hence, the larger triangle is isosceles and therefore has two equal base angles, and thus so too does the trapezoid.

share|cite|improve this answer

There is some disagreement as to whether a trapezooid must only have one pair of parallel sides. The source you cite above doesn't seem to hold this restrictive view, so a parallelogram would also be technically considered to be a trapezoid. In this case the condition on the angles is necessary to differentiate from a parallelogram.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.