1.

SIDES AB,BC AND CA OF ΔABC touches a circle at F,D and E.Prove that AF +BD+CE=1/2*perimeter of ΔABC

Answer»

ong>Step-by-step explanation:

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.


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