Slide 1

Report
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| = 90O
To Prove:
B
|BC|2 = |AB|2 + |AC|2
Construction: Draw a perpendicular from A to meet BC at D
Proof:
In the triangles ABC and ADC
B
90o
D
D
90o
90o
A
90o
A

C
A
90o
C
|  BAC| = |ADC|
Right angles
|  BCA| = |ACD|
Same angle
∆ ABC and ∆ ADC are equiangular.
 ∆ ABC and ∆ ADC are similar.

Theorem 13
| BC | | AC |

| AC | | DC |
 |AC|2 = |BC| x |DC| ………….Equation 1
© Project Maths Development Team
C
In the triangles ABC and ABD
B
B
B


90o D
90o
D
90o
90o
A

C
A
|  BAC| = |ADB|
Right angles
|  ABC| = |ABD|
Same angle
A
90o
C
∆ ABC and ∆ ABD are equiangular.
 ∆ ABC and ∆ ABD are similar.
| BC | | AB |


| AB | | BD |
Theorem 13
 |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
B
 |AC|2 + |AB|2 = |BC| {|DC| + |BD|}
But |AC|2 + |AB|2 = |BC| {|DC| + |BD|}
|BC| is common
|BC| is common
90o
90o
But DC| + |BD| = |BC|



|AC|2 + |AB|2 = |BC| x |BC|
D
A
90o
C
|AC|2 + |AB|2 = |BC|2
|BC|2 = |AB|2 + |AC|2
Q.E.D.
© Project Maths Development Team 2009

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