Answer:
SEE EXPLANATION
Step-by-step explanation:
[tex] In\:\triangle AOC \:\&\:\triangle BOD[/tex]
[tex] AO \cong OB.... (given) [/tex]
[tex] \angle AOC \cong\angle BOD [/tex]
[tex] (vertical \: \angle s) [/tex]
[tex] CO \cong OD.... (given) [/tex]
[tex] \therefore \triangle AOC \:\cong\:\triangle BOD[/tex]
[tex] (SAS \: postulate) [/tex]
So, the 3 pairs of equal parts in ∆AOC and ∆BOD are:
[tex] AC = BD[/tex]
[tex] m\angle OAC = m\angle OBD[/tex]
[tex] m\angle OCA = m\angle ODB[/tex]
