How do you prove that the diagonals and either base of an isosceles trapezoid form an isosceles triangle?

Answer 1

Consider the isosceles trapezium ABCD (going clockwise from top left) with AB parallel to CD. And let the diagonals intersect at O

Since it is isosceles, AD = BC

and <ADC = <BCD (the angles at the base BC).

Now consider triangles ADC and BCD.

AD = BC

The side BC is common

and the included angles are equal.

So the two triangles are congruent.

and therefore <ACD = <BDC

Then, in triangle ODC,

<OCD (=<ACD = <BDC) = <ODC

ie ODC is an isosceles triangle.

The triangle formed at the other base can be proven similarly, or by the fact that, because AB CD and the diagonals act as transversals, you have equal alternate angles.

Answer 2

Instead of writing it as a formal proof , which is inconvenient without a drawing and an easy way to write symbols, which I don't have at the moment, I'm going to just talk through it.

Ok, by definition an isosceles triangle has base angles which are equal to each other and sides that are congruent to each other. The base obviously has a length equal to its own length. So by SAS the triangle formed by the base, the right base angle, and the diagonal running from top right to bottom left is congruent to the triangle formed by the left side, the base and the diagonal running from the top left to the bottom right.

That's easy. What was bothering me was how to prove that the diagonals cut each other into two pairs of equal-length segments and then from there to show that they are in the proportion equal to the proportion of the top and the bottom. So this proof I just explained was step one and once that is established the rest isn't very hard.

You can show that the top triangle in the center is isosceles and the bottom center one is as well using the CPCTC theorem with the proof I explained above. There are two ways to show that the top and bottom center triangles are similar to each other - that I notice at the moment - one is that by vertical angles theorem the top and bottom angles of the X formed in the center of the triangles are of equal size and therefore the other two, splitting what is left, must be congruent to each other. The other is that by AIA theorem the base angles formed between the base and the diagonals are equal to the ones on the top and the other side of the angles formed between the top and the diagonals. You may have to stop and draw them and think about what I've just said to understand that. I think that's harder to put into words than the things I said in the proof explanation above. Now we know that the top triangle is similar to the bottom triangle. (the middle top one and the middle bottom one. Therefore the pieces that the diagonals cut each other into form two pairs of congruent lengths and since they are from similar triangles they are in the same proportion as the bases of those triangles.

That whole thing was an ACT problem. I could see it intuitively but I couldn't explain it to a student in a coherent way without first doing the proof I explained at the top above.