D is the extension line from DF‖AC to BC at F.

In the isosceles trapezoid ABCD

AC=BD

AC=DF again

∴BD=DF

It's DE⊥BC in e again

∴BD⊥DF

△BDF is an isosceles right triangle.

AD+BC= 10。

∴BF= 10

BD=DF=5√2

You are DE⊥BC

∴DE is the vertical center line of blast furnace.

∴BE=5

DE=√(5√2)？ -5? =5

2.

Intercept a little e on BC line

Such that CE=CD

△CED is an isosceles triangle

CD = CE

∴∠CDE=∠CED=( 180 -80 )/2=50

B.C.

∴∠A= 130

∠D= 100

∠ CDE = 50。

∴∠ADE=50 =∠B

∠DEB= 180 -50 = 130 =∠A

∴ Quadrilateral ABED is a parallelogram (two groups of quadrangles with equal diagonals are parallelograms)

∴AD=BE

And BC=BE+EC.

∴BC=AD+DC