In known trapezoidal ABCD, ad is parallel to BC (AD is less than BC), m and N are two waist AB, the midpoint of CD, me is parallel to an, BC intersects e, am = ne is proved

In known trapezoidal ABCD, ad is parallel to BC (AD is less than BC), m and N are two waist AB, the midpoint of CD, me is parallel to an, BC intersects e, am = ne is proved

Connecting Mn angle B = amn angle ban = BME am = MB, so △ MBE congruent △ amn gets an = me, Mn is the common edge, ∠ anm = NME, so △ amn congruent △ MNE, so am = ne

In trapezoidal ABCD, AD / / BC (AD < BC), m, n are the midpoint of two waist AB, CD, me / / an intersects BC with E

Extend an to BC extension line p
Me / / an, M is the midpoint of AB, me is the median line of triangle ABP
So: me = 1 / 2AP
Ad / / BC, n is the midpoint of CD
So: n is the midpoint of AP, an = AP / 2
So: an = me, and me / / an
So: ame is a parallelogram
So: am = ne

In the square ABCD, there is a little e, which satisfies ∠ EBC = ∠ ECB = 15 °. What triangle is Δ ade?
Please attach an explanation. Thank you

Equilateral triangle
Now I'll prove it backwards
The angle ead is 60 ° and the angle EAB is 30 ° when ade is an equilateral triangle
Since AB equals AE, the angle Abe equals the angle AEB equals 75 degrees
Then the angle EBC is equal to 15 ° which is given in the title
The proposition is proved!