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In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides.

\(AC^2=AB^2+BC^2\)

Proof

We know, △ADB ~ △ABC

Therefore, \(\tfrac{AD}{AB}=\tfrac{AB}{AC} \)(corresponding sides of similar triangles)

Or, AB^{2 }= AD × AC ……………………………..……..(1)

Also, △BDC ~△ABC

Therefore, \(\tfrac{CD}{BC}=\tfrac{BC}{AC}\) (corresponding sides of similar triangles)

Or, BC^{2}= CD × AC ……………………………………..(2)

Adding the equations (1) and (2) we get,

AB^{2 }+ BC^{2 }= AD × AC + CD × AC

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

Since, AD + CD = AC

Therefore, AC^{2} = AB^{2} + BC^{2}

Hence, the Pythagorean theorem is proved.

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