You can
put this solution on YOUR website!
Draw DF and EG ⊥ AC
DF ⊥ AC (drawn that way)
EG ⊥ AC (drawn that way)
DF ∥ EG lines perpendicular to the same line are parallel
DE ∥ FG bases of a trapezoid are parallel
FDEG is a parallelogram (both pairs of opposite sides parallel)
DF ≅ EG Opposite sides of a parallelogram
DA ≅ EC legs of isosceles trapezoid
ᐃADF and ᐃCEG are right triangles (DF and EG ∥ AC)
ᐃADF ≅ ᐃCEG hypotenuse-leg theorem
∠A ≅ ∠C corresponding parts of congruent triangles ADF and CEG
ᐃABC is isosceles base angles congruent.
Edwin
