Math, asked by sourabhsingh94995, 1 year ago

the incircle of triangle ABC touches the sides is bc CA and ab at D E and F respectively show that a f + b d + c e = a e + b f + c d equals to half perimeter of triangle ABC​

Answers

Answered by TakenName
1

Inner circle touches sides AB, BC, and CA at F, D, and E.

We know that

AB ⊥ IF,

BC ⊥ ID,

and CA ⊥ IE.

Because of inner circle I,

AE = AF,

BD = BF,

and CD = CE.

1.

∴ ΔAEI ≡ ΔAFI (RHS)

∴ ΔBDI ≡ ΔBFI (RHS)

∴ ΔCDI ≡ ΔCEI (RHS)

2.

Because of 1, we know that

AE = AF,

BD = BF,

and CD = CE.

∴ AF + BD + CE = AE + BF + CD (∵ 2)

Perimeter of ΔABC equals AB + BC + CA.

AB + BC + CA

= (AF + BF) + (BD + CD) + (CE + AE)

= (AE + AF) + (BD + BF) + (CD + CE)

3.

∴ (AF + BD + CE) + (AE + BF + CD) is the perimeter of ΔABC.

Because of 2 and 3,

both AF + BD + CE and AE + BF + CD equals to half perimeter of ΔABC.

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