If the vertical line passing through A and D as BC passes through P and Q, BP = CQ =1/2 (4 * 4 √ 2-4 √ 2) = 6 √ 2.

∠ABC=45, so AB = ∠ 2 * 6 ∠ 2 = 12. If △ABE is an isosceles triangle and △ ABC = 45, that is, △ABE is an isosceles right triangle, then EP = BP = 6 ∠ 2, ∠ AEB = 45.

2. In this question ∠ A+∠ B = 90, it can be assumed that ∠A=∠B=45, then MN is the height of a trapezoid, AB passes through the vertical lines of C and D, then DP = AP = 1/2 (664.

3. When PA+PD is the minimum, then △APD is an isosceles triangle, and if D is used to make DE perpendicular to BC in E, then △ABP=PE= 1. Because AD = 2, BC = 5, CE = 3, and CD = 5, in the right angle △CED, DE = 5 2-3 2 = AP = √17, and the area of △ APD can be expressed as AD*AB/2 or AP*AP side height /2, so the AP side height.

You can easily solve this problem by drawing a picture. After A made an AE parallel CD in E and submitted it to BC, BE=BC-CE=4- 1=3, ∠AEB=70, and because ∠B=40, in △ABE, ∠ BAE = 6544.

I have been writing for a long time. I don't know if it's right. I hope it helps you.