Proof of Pythagorean Theorem

The proof of Pythagorean Theorem in mathematics is very important.

In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).

In short it is written as: $a^2+b^2=c^2$

Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.


Then, we will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the

square EFGH, each of whose side is a, so area of the square EFGH is $a^2$

Now, we are sure that square
WXYZ = square EFGH + 4 ∆ GYF
or, $(b+c{)}^2=a^2+4\times 1/2bc$
or, $b^2+c^2+2bc=a^2+2bc$
or, $b^2+c^2=a^2$

Proof of Pythagorean Theorem using Algebra:

Given: A ∆ XYZ in which ∠XYZ = 90°.
To prove: $X{Z}^2=X{Y}^2+Y{Z}^2$

Construction: Draw YO ⊥ XZ

Proof: In ∆XOY and ∆XYZ, we have,

∠X = ∠X → common

∠XOY = ∠XYZ → each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ → by AA-similarity

⇒ XO/XY = XY/XZ

⇒ XO × XZ = XY2 ----------------- (i)
In ∆YOZ and ∆XYZ, we have,

∠Z = ∠Z → common

∠YOZ = ∠XYZ → each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ → by AA-similarity

⇒ OZ/YZ = YZ/XZ

⇒ $OZ\times XZ=Y{Z}^2$ ----------------- (ii)

From (i) and (ii) we get,

$XO\times XZ+OZ\times XZ=(X{Y}^2+Y{Z}^2)$

⇒ $(XO+OZ)\times XZ=(X{Y}^2+Y{Z}^2)$

⇒ $XZ\times XZ=(X{Y}^2+Y{Z}^2)$

⇒ $X{Z}^2=(X{Y}^2+Y{Z}^2)$

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