Draw a line through the points D and E to meet AB and F.
Consider triangles ADE and BDE.
Side DE is common to both triangles. and AE=BE and AD=BD (both given)
Therefore the 2 triangles are congruent by SSS and so m<ADE = m<BDE.
In a similar way triangles AEF and BEF are congruent and so
m <AEF = m<BEF.
So we see that the line DEF bisects both angles ADB and AEB and since both triangles ADB and AEB are isosceles then DE is perpendicular to AB.
