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: a2 + b2 = c2
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
Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF
or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c
or, b2 + c2 +
or, b2 + c2 = a2
Proof of Pythagorean Theorem using Algebra:
To prove: XZ2 = XY2 + YZ2
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 × XZ = YZ2 ----------------- (ii)From (i) and (ii) we get,
XO × XZ + OZ × XZ = (XY2 + YZ2)
⇒ (XO + OZ) × XZ = (XY2 + YZ2)
⇒ XZ × XZ = (XY2 + YZ2)
⇒ XZ 2 = (XY2 + YZ2)
Conditions for the Congruence of Triangles
Right Angle Hypotenuse Side congruence
Converse of Pythagorean Theorem
7th Grade Math Problems
8th Grade Math Practice
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