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  • In a right triangle, prove that the square of the hypotenuse is equal to the sum of squares of the other two sides.       

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

 

 
 
 
 
 

Answers (1)

Write the correct proof with figure

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)

Add (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

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Safeer PP

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