Question

F
11) In given figure , the circle of AABC, touches the sides BC,CA & AB at D,E &
Frespectively.Show that AF + BD + CE = AE + BF + CD = (perimeter of
ДАВС)
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Answer 1

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

△ABC = AB + BC + AC

= (AF + FB) + (BD + CD) + (AE + EC)

= (AF + AE) + (BF + BD) + (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 = ½ (Perimeter of △ABC)

Hence, AF + BD + CE = AE + BF + CD = ½(Perimeter of △ABC)

Hence proved.