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Pythagoras theorem

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in a right angled triangle the sum of squares of two sides is equal to the square of its hypotenuse
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In DeltaABC , /_C is an acute angle, seg AD bot seg BC . Prove AB^(2)=BC^(2)+AC^(2)-2BCxxDC by completing the following activity. Let AB=c , AC=b , AD=p , BC=a , DC=x :.BD=a-x In DeltaADB , by Pythagoras theorem, c^(2)=(a-x)^(2)+square c^(2)=a^(2)-2ax+x^(2)+square ........... (1) In DeltaADC , by Pythagoras theorem, b^(2)=p^(2)+square p^(2)=b^(2)-square .......... (2) Substituting value of p^(2) from (2) in (1) c^(2)=a^(2)-2ax+x^(2)+square c^(2)=a^(2)+b^(2)-square :.AB^(2)=BC^(2)+AC^(2)-2BCxxDC .

Without using Pythagoras's theorem, show that the points A(1, 2), B(4, 5) and C(6, 3) are the vertices of a right-angled triangle.

(Pythagoras's Theorem) Prove by vector method that in a right angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Theorems

Theorems

Which of following does not form a Pythagoras triplet ?

Pythagoras was a student of