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

Mathematics

2 answers:

sergeinik [125]2 years ago

8 0

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

Simora [160]2 years ago

6 0

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.

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