Prove That, in a Right-angled Triangle, the Square of Hypotenuse is Equal to the Sum of the Square of Remaining Two Sides - Geometry Mathematics 2

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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)

∴ AB2 = 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`

  Is there an error in this question or solution?
2017-2018 (March) Set A

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