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Question
Prove that, in a right angled triangle, the square of the hypotenuse is
equal to the sum of the squares of remaining two sides.
Solution
Given : In Δ ABC, ∠ ABC = 90o
To Prove : AC2 = AB2 + BC2
Construction : Draw seg BD perpendicular
on side AC. A-D-C.
Proof : In right angled Δ ABC,
seg BD ⊥ hypotenuse AC ...... (Construction)
∴ Δ ABC ∼ Δ ADB ∼ Δ BDC ...(Similarity of right angled triangles)
Δ ABC ∼ Δ ADB
`∴(AB)/(AD)=(BC)/(DB)=(AC)/(AB)` .....(corresponding sides)
`∴(AB)/(AD) =(AC)/(AB)`
∴ AB2 = AD × AC ....... (I)
Δ ABC ∼ Δ BDC
`(AB)/(BD) =(BC)/(DC) =(AC)/(BC)`......(corresponding sides)
`∴ (BC)/(DC) = (AC)/(BC)`
∴ BC2 = AC × DC ....... (II)
Adding (I) and (II)
`AB^2 + BC^2= AD × AC + AC × DC`
`∴ AB^2 + BC^2 = AC (AD + DC)`
`∴ AB^2 + BC^2 = AC × AC ...... (A-D-C)`
`∴ AB^2 + BC^2 = AC^2`
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