The correct option is C Both (a) and (b)
Given,
ABCD is a quadrilateral with DA⊥AB and CB⊥AB
and EF || AD
⇒EF||BC
⇒EF⊥AB
Now, in ΔDAE and ΔCBE,
∠DAE=∠CBE [each 90∘]
Also, ∠DEA=∠CEB
∴ΔDAE∼ΔCBE [by AA similarity]
⇒DEAE=CEBE⇒DE=CE ...(i)
Now, in ΔDEC, right angled at E,
DE2+CE2=DC2⇒2DE2=2CE2=DC2