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Question Number 194815 by universe last updated on 16/Jul/23

	Answered by sniper237 last updated on 16/Jul/23
	$Incenter of ∠ABC is bar{(A,a),(B,b),(C,c)} meaning aIA^→ +bIB^→ +cIC^→ =0^→ (∗) hence aI′A^→ ′+bI′B^→ ′+cI′C^→ ′=0^→ (∗∗) by projecion Afer inroducing I′ in (∗) we have (a+b+c)II^→ ′ = a(AA^→ +A′I^→ ′) + b(BB^→ ′+B′I^→ ′)+c(CC^→ ′+C′I^→ ′) = a AA^→ ′+bBB^→ ′+cCC^→ ′ applying (∗∗) so (a+b+c)II′=a.a′+b.b′+c.c′ since all following same direcion$
	$I n c e n t e r o f ∠ A B C i s b a r {(A, a), (B, b), (C, c)}$ $m e a n i n g a I \vec{A} + b I \vec{B} + c I \vec{C} = \vec{0} ()$ $h e n c e {a I}^{'} {\vec{A}}^{'} + {b I}^{'} {\vec{B}}^{'} + {c I}^{'} {\vec{C}}^{'} = \vec{0} ( ) b y p r o j e c i o n$ $A f e r i n r o d u c i n g I^{'} i n () w e h a v e$ $(a + b + c) I {\vec{I}}^{'} = a (A \vec{A} + A^{'} {\vec{I}}^{'}) + b (B {\vec{B}}^{'} + B^{'} {\vec{I}}^{'}) + c (C {\vec{C}}^{'} + C^{'} {\vec{I}}^{'})$ $= a A {\vec{A}}^{'} + b B {\vec{B}}^{'} + c C {\vec{C}}^{'} a p p l y i n g (* *)$ $s o (a + b + c) {I I}^{'} = a . a^{'} + b . b^{'} + c . c^{'} s i n c e a l l f o l l o w i n g s a m e d i r e c i o n$
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