Question

proof of Pythagoras theorem

Answer 1

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

Answer 2

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