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Question Number 209220 by alcohol last updated on 04/Jul/24

	Answered by Berbere last updated on 04/Jul/24

	$S A B, S A C & A B C c e s t c l a i r e (A S) ⊥ (A B C) & A B C r e c t a n g l e$ $\Rightarrow {S A}^{2} + {A B}^{2} = {S B}^{2}; {S A}^{2} + {A C}^{2} = {S C}^{2}; {C A}^{2} + {C B}^{2} = {A B}^{2}$ $(1) & (2) \Rightarrow {S C}^{2} - {S B}^{2} = {A C}^{2} - {A B}^{2}$ $\Rightarrow {S B}^{1} - {S C}^{2} = {C B}^{2} \Rightarrow {S B}^{2} = {C B}^{2} + {S C}^{2} \Rightarrow S B C e s t r e c t a n g l e e n C$ $(2)$ $A F ⊥ (S B)$ $A \vec{F} . C \vec{B} = (A \vec{C} + \vec{C F}) . C \vec{B} = A \vec{C} . C \vec{B} + C \vec{F} . C \vec{B} = 0 + 0 = 0$ $f i r s t A C B r e c t a n g l e (2); S C B r e c t a n g l e C \vec{F} = a S \vec{C}$ $(A F) ⊥ (C B)$ $(2) b A F \in (S A C) a n d (A F) ⊥ (S B C) \Rightarrow (A S C) ⊥ (S B C)$ $(A C) \in (S A C) & (A C) ⊥ (B C); (A C) ⊥ (S C)$ $\Rightarrow (A C) \in (A S C) & (A C) ⊥ (S B C) \Rightarrow (A S C) ⊥ (S B C)$ $(3)$ $(A F) \in (S A C) ⊥ (S B C)$ $\Rightarrow (A F) ⊥ (B C)$ $A \vec{F} = A \vec{S} + a S \vec{C}; A \vec{F} . A \vec{C} = 0 c a r l e s f a c e d u T e t r a e d e S o n t$ $o r t h o g o n a l$ $(A F) ⊥ (A C) \Rightarrow (A F) ⊥ (A B C) \Rightarrow (C F A) ⊥ (A B C)$ $(A F) ⊥ (S B C); (A F) \in (A D F) \Rightarrow (A D F) ⊥ (S B C)$ $D F \in (A D F); (S B) \in (S B C) \Rightarrow (D F) ⊥ (S B)$ $(5) m e m e m e t h o d e$
	Commented by alcohol last updated on 05/Jul/24

	$l e s c h e m a p a r d o n$
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