Question

Q. In fig., the incircle of ∆ABC, touches the sides BC, CA and AB at
D, E, F respectively. Show that
1
AF + BD + CE = AE + BF + CD =1/2(Perimeter of ∆ ABC).


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Answer 1

Answer:

Given : ΔABC circumscribes the circle

To prove that : AF + BD + CE = AE + BF + CD = 1/2 (perimeter of ΔABC)

Proof : In Δ ABC

AD = AF (tangents from same point are equal)-(1)

CF = CE ( " " " )-(2)

DE = DB (" " " )-(3)

Adding (1),(2)&(3)

AF + CE + BD = AD + CF + DE

NOW

PERIMETER OF ΔABC = AB + BC + AC

= 2 AF + 2 CE + 2 DB

= 2 ( AF + CE + BD) -(4)

THEREFORE EQUATING (1) & (4)

AF + CE + DB = 1/2 ( Perimeter of ΔABC)

Hope it helps it took quite some time typing it out

Answer 2

Answer:

Given : ΔABC circumscribes the circle

To prove that : AF + BD + CE = AE + BF + CD = 1/2 (perimeter of ΔABC)

Proof : In Δ ABC

AD = AF (tangents from same point are equal)-(1)

CF = CE ( " " " )-(2)

DE = DB (" " " )-(3)

Adding (1),(2)&(3)

AF + CE + BD = AD + CF + DE

NOW

PERIMETER OF ΔABC = AB + BC + AC

= 2 AF + 2 CE + 2 DB

= 2 ( AF + CE + BD) -(4)

THEREFORE EQUATING (1) & (4)

AF + CE + DB = 1/2 ( Perimeter of ΔABC)

Step-by-step explanation:

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