From A draw AG // BC cuts CD at WO
We have: Trapezoid ABCD (assumption)
⇒ AB // CD
⇒ AB // GC (because G ∈ CD)
Considering quadrilateral ABCG, there are:
AB // GC (proven above)
AG // BC (assumption)
⇒ Quadrilateral ABCG is a parallelogram
⇒ AB = GC = 40 cm
AG = BC = 50 cm
We have: DG = CD - GC (because G ∈ CD)
DG = 80 - 40
⇒ DG = 40(cm)
Considering ΔAGD, there are:
AG2=AD2+DG2AG2=AD2+DG2
502=302+402⇒502=302+402
⇒502=900+1600⇒502=900+1600
502=2500⇒502=2500
⇒502=502⇒502=502
⇒ AGD square at D
⇒ Trapezoid ABCD is a square trapezoid
