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

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

 

 

Answers (1)

A right triangle ABC in which \angle B = 90 \degree \\\\

 To prove : (hypotenuse)^2 = (Base )^2 + ( perpendicular)^2 \\\\ AC ^2 = AB ^ 2 + BC ^2 \\\\ construction \: \: BD \perp AC \\\\

proof : in \: \Delta ABC \: \: and \: \: \Delta ADB\\\\ \angle ADB = \angle ABC \\\\ \angle A = \angle A \\\\

so, by AA similarity criterion 

\Delta ABC \: \: \sim \: \: \Delta ADB\\\\

= \frac{AD }{AB } = \frac{AB }{AC }  [in similar triangles corresponding sides are ]

 

AB^ 2 = AD . AC ---(1) \\\\ In \: \: \Delta BDC \: \: \: \: and\ \ \Delta ABC \\\\ \angle CDB = \angle ABC \\\\ \angle C = \angle C \\\\

so by AA similarity 

\Delta BDC \sim \Delta ABC

=\frac{DC }{BC} = \frac{BC}{AC} ----(2) \\\\

adding equation (1) and (2) 

AB^2 + BC ^ 2= AC ( AD + DC ) \\\\ AB^2 + BC^2 = AC . AC \\\\ AB ^ 2 + BC ^2 = AC^ 2 \\\\ hence \: \: AC ^ 2= AB ^ 2 + BC^

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