answer = Isosceles trapezoid

From the three types of trapezoid:

1. isosceles trapezoid

2. Right trapezoid

3. Scalene trapezoid (i.e. neither 1. nor 2.)

cases/ types 2. and 3. can be eliminated by contradiction

(technically, 2. is a partial case of 3.)

To eliminate 3, assume 3. is a possible case, and consider the altitude (with length, say, h) from vertices B and C to the base AD, and call the three line segments from AD to be a, b and c.

Then, the given one, AB+BD=AC+CD can be writen as

sqrt(a^2+h^2)+sqrt((c+b)^2+h^2) = sqrt((a+b)^2+h^2)+sqrt(c^2+h^2)

When we try to solve it for "h" then we get h=0, which is the contradiction.

(the partial case 2. corresponds to a=0, or c=0)

https://www.wolframalpha.com/input?i=solve+sqrt%28a%5E2%2Bh%5E2%29%2Bsqrt%28%28c%2Bb%29%5E2%2Bh%5E2%29+%3D+sqrt%28%28a%2Bb%29%5E2%2Bh%5E2%29%2Bsqrt%28c%5E2%2Bh%5E2%29+for+h

similar to the above wolframalpha.com-link where the solution for "h" is "0",

we can ask for a solution for "a", for example:

https://www.wolframalpha.com/input?i=solve+%28a%5E2%2Bh%5E2%29%282*b*c%2Bb%5E2%29%2B2%28c*b-a*b%29sqrt%28a%5E2%2Bh%5E2%29sqrt%28%28c%2Bb%29%5E2%2Bh%5E2%29+%2B%28c*b-a*b%29%5E2+%3D+%282*a*b%2Bb%5E2%29%28c%5E2%2Bh%5E2%29+for+a

and, in such case, solutions are "a=c" and "a=-b-c", that is, again we confirm case 1. 'isosceles trapezoid' - certainly, the whole thing can be derived by hand - with pen and paper.

Dear Steftcho P. Dokov I would like to share with you my solution to this problem. You might find this interesting.

Based on the conditions of the problem, we can conclude that points B and C lie on an ellipse with focuses A and D. Since BC is parallel to AD, points B and C are symmetrical about the vertical semi-axis of the ellipse. Therefore the trapezoid is isosceles.

Liudmyla Hetmanenko, Sergey E. Gladun, of course, your purely geometric proofs are much more suitable for such geometric problems than some clumsy algebraic manipulations. I like what you did.

Steftcho P. Dokov Liudmyla Hetmanenko Thank you very much.

Steftcho P. Dokov Sergey E. Gladun I'm glad that such problems are interesting not only to me. I am very grateful to you for your time and such interesting solutions.