Question
In a $\triangle\text{ ABC},\text{ BD}\perp\text{AC}$ and $\text{ CE}\perp\text{AB}.$ If BD and CE intersect O, prove that $\angle\text{BOC}=180^\circ-\text{A}.$
✓
Answer
In quadrilateral AEOD$\angle\text{A}+\angle\text{AEO}+\angle\text{EOD}+\angle\text{ADO}=360^\circ$$\Rightarrow\angle\text{A}+90^\circ+90^\circ+\angle\text{EOD}=360^\circ$
$\Rightarrow\angle\text{A}+\angle\text{BOC}=180^\circ$ $[\therefore\angle\text{EOD}=\angle\text{BOC}$ vertically opposite angles$]$
$\Rightarrow\angle\text{BOC}=180^\circ-\angle\text{A}$
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