Question

ABCD is parallelogram. The angle bisectors of ∠A and ∠D intersect at O. Find the measures of ∠AOD.

Answer 1

$$\begin{gathered} \mathrm{In}\mathrm{parallelogram}\mathrm{ABCD}, \\ \angle \mathrm{A}+\angle \mathrm{D}=180°\left(\mathrm{Adjacent}\mathrm{angles}\mathrm{of}\mathrm{a}\mathrm{parallelogram}\mathrm{are}\mathrm{supplementary}\right) \\ \Rightarrow \frac{1}{2}\angle \mathrm{A}+\frac{1}{2}\angle \mathrm{D}=\frac{180°}{2} \\ \Rightarrow \angle \mathrm{OAD}+\angle \mathrm{ODA}=90°.....\left(\mathrm{i}\right)\left(\mathrm{As},\mathrm{OA}\mathrm{and}\mathrm{OD}\mathrm{are}\mathrm{angle}\mathrm{bisectors}\mathrm{of}\angle \mathrm{A}\mathrm{and}\angle \mathrm{D},\mathrm{respectively}\right) \\ \mathrm{Now},\mathrm{in}∆\mathrm{AOD}, \\ \angle \mathrm{AOD}+\angle \mathrm{OAD}+\angle \mathrm{ODA}=180°\left(\mathrm{Angle}\mathrm{sum}\mathrm{property}\mathrm{of}\mathrm{triangle}\right) \\ \Rightarrow \angle \mathrm{AOD}+90°=180°\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \angle \mathrm{AOD}=180°-90° \\ \therefore \angle \mathrm{AOD}=90° \end{gathered}$$