Question
$AC, AD$ and $AE$ are joined. Find. $\angle \text{FAB}+\angle \text{ABC}+\angle \text{BCD}+\angle \text{CDE} +\angle \text{DEF}+\angle \text{EFA}$
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Answer
We know that sum of the angles of a triangle is $180^\circ $
Therefore in $\triangle \text{ABC},$ we have $\angle \text{CAB}+\angle \text{ABC}+\angle \text{BCA}=180^\circ...(\text{i})$ In $\triangle \text{ACD},$ we have $\angle \text{DAC}+\angle \text{ACD}+\angle \text{CDA}=180^\circ...(\text{ii})$ In $\triangle \text{ADE},$ we have $\angle \text{EAD}+\angle \text{ADE}+\angle \text{DEA}=180^\circ...(\text{iii})$ In, $\triangle \text{AEF},$ we have $\angle \text{FAE}+\angle \text{AEF}+\angle \text{EFA}=180^\circ...(\text{iv})$ Adding $(i), (ii), (iii), (iv)$
we get $\angle \text{CAB}+\angle \text{ABC}+\angle \text{BCA}+\angle \text{DAC}+\angle \text{ACD}$
$+\angle \text{CDA}+\angle \text{AD}+\angle \text{ADE}+\angle \text{DEA}+\angle \text{FAE}$
$+\angle \text{AEF}+\angle \text{EFA}=720^\circ$ Therefore $\angle \text{FAB}+\angle \text{ABC}+\angle \text{BCD}$
$+\angle \text{CDE}+\angle \text{DEF}+\angle \text{EFA}=720^\circ$
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