Question

Given: BD = BF DE ⊥ BC , FK ⊥ AB Prove: ED ≅ FK

Answer 1

This is the picture I am not very good at proofs but hopefully someone can solve it.

Answer 2

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.