#### 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.