Question
In the adjoining figure, show that$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}=360^\circ.$
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Answer
In $\triangle\text{ACE},$ we have,$\angle\text{A}+\angle\text{C}+\angle\text{E}=180^\circ\ ...(\text{i)}$
In $\triangle\text{BDF},$ we have,$\angle\text{B}+\angle\text{D}+\angle\text{F}=180^\circ\ ...(\text{ii)}$
Adding both sides of (i) and (ii), we get,$\angle\text{A}+\angle\text{C}+\angle\text{E}+\angle\text{B}\\+\angle\text{D}+\angle\text{F}=180^\circ+180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}=360^\circ.$
In $\triangle\text{BDF},$ we have,$\angle\text{B}+\angle\text{D}+\angle\text{F}=180^\circ\ ...(\text{ii)}$
Adding both sides of (i) and (ii), we get,$\angle\text{A}+\angle\text{C}+\angle\text{E}+\angle\text{B}\\+\angle\text{D}+\angle\text{F}=180^\circ+180^\circ$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}=360^\circ.$
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