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