Hi all,

could y’all me solve this proof? i’ve shown some of my work, although i’m pretty sure it’s (pic is attached below)

i’ll mark !

Arc BC = arc DC because their subtended chords BC and DC are equal

Step-by-step explanation:

∵ CE is a diameter of circle A(chord passing through the center of the circle)

∵ CE ⊥ to the chord BD and intersect it at F

∴ BF = FD ⇒ fact in the circle

In the 2 Δ CFB and CFD

∵ BF = DF

∵ m∠BFC = m∠DFC = 90°

∵ CF is a common side

∴ The two triangles are congruent

∴ BC = DC

In the circle if the two chords are equal in lengths then their subtended arcs are equal in lengths and measures

∴ Arc BC = arc DC

your answer is the frist one

hop i