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
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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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