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Published byAbby Wears Modified over 2 years ago

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Theorem 14: [Theorem of Pythagoras] In a right-angled triangle the square of the hypotenuse is the sum of the squares of the other two sides. USE THE FORWARD AND THE BACK ARROWS ON THE KEYBOARD TO VIEW AND REWIND PROOF. | Given: Triangle ABC with | BAC| = 90 O To Prove: |BC| 2 = |AB| 2 + |AC| 2 © Project Maths Development Team Proof: In the triangles ABC and ADC 90 o A C B Construction: Draw a perpendicular from A to meet BC at D 90 o D AC B A D C | BAC| = | ADC| Right angles | BCA| = | ACD| Same angle ∆ ABC and ∆ ADC are equiangular. ∆ ABC and ∆ ADC are similar. Theorem 13 |AC| 2 = |BC| x |DC| ………….Equation 1

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© Project Maths Development Team 2009 In the triangles ABC and ABD 90 o A C B D AC B | BAC| = | ADB| Right angles | ABC| = | ABD| Same angle ∆ ABC and ∆ ABD are equiangular. ∆ ABC and ∆ ABD are similar. Theorem 13 |AB| 2 = |BC| x |BD| ………….Equation 2 90 o A D B Adding Equation 1 and Equation 2 |AC| 2 = |BC| x |DC| …Equation 1 |AB| 2 = |BC| x |BD| … Equation 2 |AC| 2 + |AB| 2 = |BC| x |DC| + |BC| x |BD|

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But |AC| 2 + |AB| 2 = |BC| {|DC| + |BD|} |BC| is common But DC| + |BD| = |BC| |AC| 2 + |AB| 2 = |BC| x |BC| |AC| 2 + |AB| 2 = |BC| 2 |BC| 2 = |AB| 2 + |AC| 2 Q.E.D. © Project Maths Development Team 2009 |AC| 2 + |AB| 2 = |BC| {|DC| + |BD|} |BC| is common 90 o A C B D

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