I need help with this proof!

∠B ≅ ∠F ⇒ proved down

Step-by-step explanation:

In the two right triangles, if the hypotenuse and leg of the 1st right Δ ≅ the hypotenuse and leg of the 2nd right Δ, then the two triangles are congruent

Let us use this fact to solve the question

→ In Δs BCD and FED

∵ ∠C and ∠E are right angles

∴ Δs BCD and FED are right triangles ⇒ (1)

∵ D is the mid-point of CE

→ That means point D divides CE into 2 equal parts CD and ED

∴ CD = ED ⇒ (2) legs

∵ BD and DF are the opposite sides to the right angles

∴ BD and DF are the hypotenuses of the triangles

∵ BD ≅ FD ⇒ (3) hypotenuses

→ From (1), (2), (3), and the fact above

∴ Δ BCD ≅ ΔFED ⇒ by HL postulate of congruency

→ As a result of congruency

∴ BC ≅ FE

∴ ∠BDC ≅ ∠FDE

∴ ∠B ≅ ∠F ⇒ proved

other answer is correct. i had used the radius of the radius somehow. a website: