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Question
In Fig. the incircle of ΔABC touches the sides BC, CA, and AB at D, E respectively. Show that: AF + BD + CE = AE + BF + CD = `1/2`( Perimeter of ΔABC)
Solution
Since lengths of the tangents drawn from an exterior point to a circle are equal.
∴ AF = AE ...(i)
BD = BF ...(ii)
and CE = CD ...(iii)
Adding (i), (ii) and (iii), we get
AF + BD + CE = AE + BF + CD
Now, Perimeter of Δ ABC = AB + BC + AC
= (AF + FB) + (BD + CD) + (AE + EC)
= (AF + AE) + (BD + BF) + (CD + CE)
= 2AF + 2BD + 2CE
= 2( AF + BD + CE) ....(From (i), (ii) and (iii), we get AE = AF, BD = BF, and CD = CE)
∴ AF + BD + CE = `1/2`( Perimeter of ΔABC)
∴ AF + BD + CE = AE + BF + CD = `1/2`( Perimeter of ΔABC)
Hence proved.
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