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Pythagoras Theorem

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theorem

Theorem1: In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Given: A right triangle ABC right angled at B.
Construction: Draw BD ⊥ AC
To prove: `"AC"^2`= `"AB"^2+"BC"^2`
Proof: In ∆BAD and ∆BAC
∠A= ∠A (common)
∠BDA= ∠ABC= 90°
∆BAD ∼ ∆CAB (Angle-angle similarity)


So, `"BA"/"AC" = "AD"/"BA"` (Corresponding parts of Congruent triangles)


`"BA"^2 = "AC"xx "AD"` .....eq1
Again, In ∆BCD and ∆BAC
∠C= ∠C (common)
and ∠BDC= ∠ABC= 90°
Therefore, ∆BCD ∼ ∆ACB (Angle-Angle similarity)


`"BC"/"AC" = "CD"/"BC"` (Corresponding parts of Congruent triangles)


`"BC"^2= "AC"xx "CD"` ......eq2
Add eq1 and eq2
`"BA"^2 + "BC"^2 = ("AC"xx "AD")+ ("AC"xx "CD")`
                         = `"AC" ("AD"+"CD")`
                        = `"AC" xx "AC"`
`"BA"^2 + "BC"^2 = "AC"^2`
Hence proved.

Theorem2:  In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Given: A triangle ABC such that `"AC"^2= "AB"^2+"BC"^2`
To prove: ∠B= 90°
Construction: Draw a triangle PQR such that PQ=AB, QR=BC and ∠Q=90°
Proof: In ∆PQR,
`"PR"^2= "PQ"^2+"QR"^2` (Pythagoras theorem)
`"PR"^2= "AB"^2 + "BC"^2` (by construction) .....eq1
`"AC"^2= "AB"^2 + "BC"^2` (Given) .....eq2
From eq1 and eq2
Therefore, `"PR"^2= "AC"^2`
PR= AC
Now, In ∆ABC and ∆PQR,
AB= PQ (by construction)
BC= QR (by construction)
and AC= PR (proved above)
Therefore, ∆ABC ≅ ∆PQR (Side-Side-Side rule)
So, ∠B=∠Q= 90°  (Corresponding parts of Congruent triangles)
Therefore, ∠B= 90°
Hence proved.

description

  • Pythagoras Theorem : In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
  • In a right-angled triangle, the square on the hypotenuse is equal to the sum of the squares on the remaining two sides.
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