Proof On Pythagorean Theorem
In this section you will learn the Proof On Pythagorean Theorem.
Theorem 1 : In a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Proof On Pythagorean Theorem
Given : A right angled triangle ABC in which ∠B = 90 ^{0}
Prove that : AC ^{2} = AB ^{2} + BC ^{2}
Construction : From B draw BD ⊥ AC.



1) ∠ABC = 90^{0}  1) Given  
2) BD ⊥ AC  2) By construction  
3) ∠ADB = 90^{0}  3) By definition of perpendicular  
4) ∠ADB = ∠ABC  4) Each of 9090^{0}  
5) ∠A = ∠A  5) Reflexive  
6) ΔADB ~ ΔABC  6) By AA postulate  
7) AD/AB = AB/AC  7) By basic proportionality theorem  
8) AB^{2} = AD x AC  8) By cross multiplication  
9) ∠CDB = ∠ABC  9) Each 90^{0}  
10) ∠C = ∠C  10) Reflexive  
11) ΔBDC ~ ΔABC  11) By AA postulate  
12) DC/BC = BC/AC  12) By basic proportionality theorem  
13) BC^{2} = DC x AC  13) By cross multiplication  
14) AB^{2} + BC^{2} = AD x AC + AC x DC  14) By adding (8) and (13)  
15) AB^{2} + BC^{2} = AC(AD + DC)  15) By distributive property  
16) AB^{2} + BC^{2}= AC^{2}  16) As AC = AD + DC and by substitution  
Application based on Proof On Pythagorean Theorem.
Theorem 2 : ΔABC is an obtuse triangle, obtuse angled at B. If AD ⊥ CB, prove that AC^{2} = AB^{2} + BC^{2} + 2 BC x BD.
Given : ΔABC is an obtuse triangle, obtuse angled at B. AD ⊥ CB.
Prove that : AC ^{2} = AB ^{2} + BC ^{2} + 2 BC x BD
Proof :
In ΔADB,
AB ^{2} = AD ^{2} + DB ^{2} ( Pythagorean theorem)
In ΔADC,
AC ^{2} = AD ^{2} + DC ^{2}
AC ^{2} = AD ^{2} + (DB + BC) ^{2} ( since DC = DB + BC)
AC ^{2} = AD ^{2} + DB ^{2} +BC ^{2} + 2BD x BC [ (a +b) ^{2} = a ^{2} + b ^{2} + 2ab > identity]
AC ^{2} = (AD ^{2} + DB ^{2} ) + BC ^{2} + 2BC x BD
AC ^{2} = AB ^{2} + BC ^{2} + 2BC x BD [ from above ]
This theorem is known as Apollonius theorem.
Statements of some useful theorems
Theorem 1 : Prove that in any triangle, the sum of the squares of any two sides is equal to the twice the square of half of the third side together with twice the square of the median which bisects the third side.
AB ^{2} + AC ^{2} = 2 ( AD ^{2} + BD ^{2} )
Theorem 2 : ∠B of ΔABC is an acute angle and AD ⊥ BC.
AC ^{2} = AB ^{2} + BC ^{2}  2 BC x BD
Pythagorean Theorem
• Introduction of Pythagorean Theorem
• Converse of Pythagorean Theorem
• Pythagorean Triples
• Application On Pythagorean Theorem
• Proof on Pythagorean Theorem