Please answer this

Dear student,
$$\begin{gathered} 1) In ∆BOC and ∆AOD \\ \angle B=\angle A --[each 90°] \\ BC=AD --[given] \\ \angle BOC=\angle AOD --[vert. opp. \angle s] \\ \therefore ∆BOC\cong ∆AOD --[by ASA] \\ \Rightarrow OB=OA --[CPCT] \end{gathered}$$
For remaining question kindly post in separate thread for meaningful help.
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