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**© Project Maths Development Team**

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 |ÐBAC| = 90o 90o A C B To Prove: |BC|2 = |AB|2 + |AC|2 Construction: Draw a perpendicular from A to meet BC at D Proof: In the triangles ABC and ADC 90o D 90o A C B 90o A D C | BAC| = |ADC| Right angles | BCA| = |ACD| Same angle ∆ ABC and ∆ ADC are equiangular. Þ ∆ ABC and ∆ ADC are similar. Theorem 13 Þ |AC|2 = |BC| x |DC| ………….Equation 1 © Project Maths Development Team

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**© Project Maths Development Team 2009**

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

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**© Project Maths Development Team 2009**

B D Þ |AC|2 + |AB|2 = |BC| {|DC| + |BD|} |BC| is common 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

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Determination of an Angle || A B C D MP ON Given triangles ∆ABC and ∆ADC, having AB=AC=AD in a square □ MNOP. Line N C = C O, and BD is parallel to NO.

Determination of an Angle || A B C D MP ON Given triangles ∆ABC and ∆ADC, having AB=AC=AD in a square □ MNOP. Line N C = C O, and BD is parallel to NO.

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