In the six cornered figure, AC, AD and AE are joined. Find ∠FAB + ∠ABC + ∠BCD + ∠CDE + ∠DEF + ∠EFA.

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Solution

$We have to find ∠ FAB + ∠ ABC + ∠ BCD + ∠ CDE + ∠ DEF + ∠ EFA . . . . . . (i) From the figure, we have : ∠ FAB = ∠ FAE + ∠ EAD + ∠ DAC + ∠ CAB ∠ BCD = ∠ ACB + ∠ ACD ∠ CDE = ∠ ADC + ∠ ADE ∠ DEF = ∠ AED + ∠ AEF Putting the values of ∠ FAB, ∠ BCD, ∠ CDE, ∠ DEF in equation (i) : (∠ FAE + ∠ EAD + ∠ DAC + ∠ CAB) + ∠ ABC + (∠ ACB + ∠ ACD) + (∠ ADC + ∠ ADE) + (∠ AED + ∠ AEF) + ∠ EFA \Rightarrow (∠ ABC + ∠ ACB + ∠ CAB) + (∠ FAE + ∠ AEF + ∠ EFA) + (∠ FAE + ∠ AEF + ∠ EFA) + (∠ ADC + ∠ ACD + ∠ DAC) . . . . . (ii) We know that the sum of the three angles of a triangle is equal to 180 ° . Hence we can say the following : ∠ ABC + ∠ ACB + ∠ CAB = 180 ° (angles of △ ABC) ∠ FAE + ∠ AEF + ∠ EFA = 180 ° (angles of △ AFE) ∠ AED + ∠ ADE + ∠ EAD = 180 ° (angles of △ AED) ∠ ADC + ∠ ACD + ∠ DAC = 180 ° (angles of △ ADC) Putting these values in equation (ii) : 180 ° + 180 ° + 180 ° + 180 ° Hence, the sum of the given angles is 720 °$

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