Question

Line segments AB and CD intersect at O such that AC || DB. If ∠CAB = 35° and ∠CDB = 55°, find ∠BOD.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}\mathrm{AC}\parallel \mathrm{BD}\mathrm{and}\mathrm{AB}\mathrm{cuts}\mathrm{AC}\mathrm{and}\mathrm{BD}\mathrm{at}\mathrm{A}\mathrm{and}\mathrm{B},\mathrm{respectively}. \\ \therefore \angle \mathrm{CAB}=\angle \mathrm{DBA}(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}) \\ \Rightarrow \angle \mathrm{DBA}=35° \\ \mathrm{We}\mathrm{also}\mathrm{know}\mathrm{that}\mathrm{the}\mathrm{sum}\mathrm{of}\mathrm{all}\mathrm{three}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{is}180°. \\ \mathrm{Hence},\mathrm{for}△\mathrm{OBD},\mathrm{we}\mathrm{can}\mathrm{say}\mathrm{that}: \\ \angle \mathrm{DBO}+\angle \mathrm{ODB}+\angle \mathrm{BOD}=180° \\ \Rightarrow 35°+55°+\angle \mathrm{BOD}=180°(\because \angle \mathrm{DBO}=\angle \mathrm{DBA}\mathrm{and}\angle \mathrm{ODB}=\angle \mathrm{CDB}) \\ \angle \mathrm{BOD}=180°-90° \\ =\angle \mathrm{BOD}=90° \end{gathered}$$