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Theorem(NIB) 4.9

# Theorem(NIB) 4.9 - Theorem(NIB 4.9 The"Similar...

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Also, m( A) = 90 ° -- x ° . Also, by definition of "altitude", B H F C and B H F A are right triangles with right angles at vertices F C and F A , resp. By definition of "altitude", A F B B and A F C C are right triangles with right angles at vertices F B and F C , resp., and with A is common to both triangles. A B F B and A C F C are complements of A . A C F C 2245 A B F B . m( A C F C ) = x ° . x ° x ° Let x ° = m( A B F B ) Proof: The proofs of statements (2) and (3) are similar to the proof of statement (1) . Only the proofs of statements (1) and (4) are presented here. Proof of (1): AF A bisects F B F A F C and A 2245 F C F A B 2245 F B F A C Let ABC be an acute triangle with Orthocenter H and with feet F A , F B , and F C of the altitudes from A, B, and C, resp. [ To prove that A F A bisects F B F A F C , we need to show that m( F C F A A) = m( F B F A A). ] Theorem (NIB) 4.9, The "Similar Sub-triangles" Theorem: In an acute triangle, ABC, (using the M A -- F A -- Q A notation): (1) AF A bisects F B F

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