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Prove that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the square of remaining two sides

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#### Solution

Draw perpendicular BD from the vertex B to the side AC. A - D

In right-angled `triangle ABC`

seg BD ⊥ hypotenuse AC.

∴ by the similarity in right-angled triangles

`triangle ABC ~ triangle ADB ~ triangle BDC`

Now, `triangle ABC ~ triangle ADB`

`:. (AB)/(AD) = (AC)/(AB) ` ......(c.s.s.t)

∴ AB^{2} = AC x AD ......(1)

Also, `triangleABC ~ triangleBDC`

`:. (BC)/(DC) = (AC)/(BC)` ......(c.s.s.t)

`:. BC^2 = AC xx DC` ......(2)

From (1) and (2),

`AB^2 + BC^2 = AC xx AD + AC xx DC`

`= AC xx (AD + DC)`

`= AC xx AC` ......(A-D-C)

`:. AB^2 + BC^2 = AC^2`

`i.e, AC^2 = AB^2 + BC^2`

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