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## Answers

Given: We are given a right triangle ABC right angled at B.

To prove: AC²=AB²+BC²

Construction: Draw BD parallel to AC.

Proof: In triangles ADB and ABC, we have

Angle ADB = Angle ABC (Each 90⁰)

and, Angle A = Angle A (Common)

So, ΔABD is similar to ΔABC (AA similarity criterion)

⇒ AD/AB = AB/AC (In similar triangles corresponding sides are proportional)

⇒ AB²=AD*AC ---(1)

In triangle BDC and ABC, we have

Angle CDB = Angle ABC (Each 90⁰)

Angle C = angle C (Common)

So, ΔBDC is similar to ΔABC (In simialr triangles corresponding sides are proportional)

⇒DC/BC=BC/AC

⇒BC²=AC*DC ---(2)

Adding equations 1 and 2, we get

AB²+BC²=AD*AC+AC*DC

⇒AB²+BC²=AC(AD+DC)

⇒AB²+BC²=AC²

Hence, AC²=AB²+BC²

**Pythagoras Theorem**states that

**''In a right angled triangle the square of the hypotenuse is equal to the sum of the squares of the other two sides.''**

**Given**: A right angled triangle ABC right angled at B.

**To prove**: AC² = AB²+BC²

**Construction**: Draw BD parallel to AC.

**Proof**: In triangles ADB and ABC, we have

Angle ADB = Angle ABC = 90⁰

and, Angle A = Angle A (Common)

Therefore, ΔABD ≈ ΔABC (AA similarity)

⇒ AD/AB = AB/AC (corresponding sides are proportional in similar Δs)

⇒

**AB² = AD.AC**.........(1)

In triangle BDC and ABC, we have

Angle CDB = Angle ABC = 90⁰

Angle C = Angle C (Common)

Therefore, ΔBDC ≈ ΔABC (corresponding sides are proportional in similar Δs)

⇒ DC/BC = BC/AC

⇒

**BC² = AC.DC**...........(2)

Adding (1) and (2),

⇒ AB² + BC² = AC.DC + AD.AC

⇒ AB² + BC² = AC (DC + AD)

⇒

**AB² + BC² = AC²**

Therefore,

**AC² = AB² + BC²**

**Hence proved Pythagoras Theorem**