In rectangle ABCD, AF bisects angle BAD and intersects DC at point F, angle CAF=15 degrees. Find the measure of angle DOF.
Hi !
Since $\displaystyle \widehat{BAF}=\widehat{FAD}$ and $\displaystyle ABCD$ rectangle then $\displaystyle \widehat{FAD}=45^{\circ}$. And $\displaystyle \widehat{AFD}=45^{\circ}$ thus $\displaystyle AD=AF$.
But $\displaystyle \widehat{CAF}=15^{\circ}$ so $\displaystyle \widehat{BAC}=30^{\circ}$ and $\displaystyle \widehat{ODF}=30^{\circ}$ too.
Notice $\displaystyle \frac{AD}{DC}=\tan(30)=\frac{1}{\sqrt{3}}$ that's why $\displaystyle OD^2=\left(\frac{AD}{2}\right)^2+\left(\frac{DC}{2 }\right)^2=\frac{AD^2}{4}+\frac{3AD^2}{4}=AD^2\Rig htarrow OD=AD$.
Finally $\displaystyle OD=DF$ and $\displaystyle \widehat{ODF}=30^{\circ}$ therefore $\displaystyle \widehat{DOF}=75^{\circ}$.