Answer:
Side ED ≅ FK
Step-by-step explanation:
The given parameters are BD = BF, DE ⊥ BC and FK ⊥ AB
We have to prove ED ≅ FK
In the given triangles ΔBDE and ΔBKF
BD = BF
∠B is common and ∠BDE ≅ ∠KFB = 90°
Since these triangles fulfill the property of congruence AAS (One side and two adjacent angles are equal) therefore ΔBDE ≅ ΔBKF
Therefore ED ≅ FK
Hence proved.
