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  • Prove that in a right triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides

Prove that in a right-triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.

 

 

 

 
 
 
 
 

Answers (1)

We have \bigtriangleup AD\!B \sim \bigtriangleup ABC   (By AA similarity)

Therefore 

\frac{AD}{AB}=\frac{AB}{AC}

(in similar triangles corresponding sides are proportional )

AB^2=AD \times AC \;\;\;\;\;\; -(1)

Also, \bigtriangleup BDC \sim \bigtriangleup ABC

therefore, 

\frac{C\!D}{BC}=\frac{BC}{AC}

or BC^2= C\!D \times AC \;\;\;\;\;\;\; -(2)

Adding eq (1) and (2), we get 

AB^2+BC^2=AD\times AC + CD \times AC

AB^2+BC^2=AC(AD+CD)

AB^2+BC^2=AC\cdot AC   (from figure AD+C\!D=AC)

Therefore, AC^2=AB^2+BC^2

 

 

 

 

Posted by

Ravindra Pindel

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