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 = 900
Prove that : AC2 = AB2 + BC2
Construction : From B draw BD ⊥ AC.
1) ∠ABC = 900
2) BD ⊥ AC
2) By construction
3) ∠ADB = 900
3) By definition of perpendicular
4) ∠ADB = ∠ABC
4) Each of 90900
5) ∠A = ∠A
6) ΔADB ~ ΔABC
6) By AA postulate
7) AD/AB = AB/AC
7) By basic proportionality theorem
8) AB2 = AD x AC
8) By cross multiplication
9) ∠CDB = ∠ABC
9) Each 900
10) ∠C = ∠C
11) ΔBDC ~ ΔABC
11) By AA postulate
12) DC/BC = BC/AC
12) By basic proportionality theorem
13) BC2 = DC x AC
13) By cross multiplication
14) AB2 + BC2 = AD x AC + AC x DC
14) By adding (8) and (13)
15) AB2 + BC2 = AC(AD + DC)
15) By distributive property
16) AB2 + BC2= AC2
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 AC2 = AB2 + BC2 + 2 BC x BD.
Given : ΔABC is an obtuse triangle, obtuse angled at B. AD ⊥ CB.
Prove that : AC2 = AB2 + BC2 + 2 BC x BD
AB2 = AD2 + DB2 ( Pythagorean theorem)
AC2 = AD2 + DC2
AC2 = AD2 + (DB + BC) 2 ( since DC = DB + BC)
AC2 = AD2 + DB2 +BC2 + 2BD x BC [ (a +b) 2 = a2 + b2 + 2ab ---> identity]
AC2 = (AD2 + DB2) + BC2 + 2BC x BD
AC2 = AB2 + BC2 + 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.
AB2 + AC2 = 2 ( AD2 + BD2)
Theorem 2 : ∠B of ΔABC is an acute angle and AD ⊥ BC.