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Pythagorean-Thm-Pf

# Pythagorean-Thm-Pf - ° m ∠ A m ∠ B = 90 ° since ABC...

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BC AB = BD CB by triangle similarity . (BC) (CB) = (AB) (BD) . a 2 = c y . Q E D a 2 + b 2 = c 2 . a 2 + b 2 = c y + c x = = c ( x + y ) = c 2 . a y m( A) + m( B) = 90 ° since ABC has an angle sum of 180 ° . m( BCD) + m( B) = 90 ° since BCD has an angle sum of 180 ° . m( BCD) = m( A) , since both equal 90 ° - m( B) . Also, m( CBD) = m( CBA) , since both equal m( B) . ABC CBD by A-A(-A) similarity. AC AB = AD AC by triangle similarity . (AC) (AC) = (AB) (AD) . b 2 = c x . ABC ACD by A-A(-A) similarity. Also, m( CAD) = m( CAB) , since both equal m( A) . m( ACD) = m( B) , since both equal 90 ° - m( A) . m( A) + m( ACD) = 90 ° since ACD has an angle sum of 180
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Unformatted text preview: ° . m( ∠ A) + m( ∠ B) = 90 ° since ABC has an angle sum of 180 ° . b x Let x = AD and let y = DB . Then, c = x + y . b a c y x a b By the (NIB) "Drop a Perpendicular" Theorem, there exists a point D on AB such that CD ⊥ AB . An argument involving the Exterior Angle Theorem proves that A - D - B. [ NTS: c 2 = a 2 + b 2 ] Proof: Given: Theorem 4.4.8 (The Pythagorean Theorem): In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. a b c D D D C B A C B A C B A C A B C...
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