Question

Prove that in a right angled triangle square of the hypotenuse is equal bto sum of the squares of other two sides

Answer 1

Given: A right triangle ABC right angled at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof:

In Δ ADB and Δ ABC,

∠ ADB = ∠ ABC (each 90°)

∠ BAD = ∠ CAB (common)



Δ ADB ~ Δ ABC (By AA similarity criterion)

Now, AD/AB = AB/AC (corresponding sides are proportional)

AB2 = AD × AC … (i)

Similarly, Δ BDC ~ Δ ABC

BC2 = CD × AC … (ii)

Adding (i) and (ii)

AB2 + BC2 = (AD × AC) + (CD × AC)

AB2 + BC2 = AC × (AD + CD)

AB2 + BC2 = AC2

Hence Proved.

Answer 2

Answer:

Given :

A right triangle ABC right angled at B.

To prove :

AC² = AB² + BC²

Construction :

Draw BD ⊥ AC

Proof :

In Δ ADB and Δ ABC

∠ A = ∠ A    [ Common angle ]

∠ ADB = ∠ ABC   [ Both are 90° ]

∴  Δ  ADB  Similar to Δ ABC   [ By AA similarity ]

So , AD / AB = AB / AC   [ Sides are proportional ]

= > AB² = AD . AC  ... ( i )

Now in Δ BDC and Δ ABC

∠ C = ∠ C    [ Common angle ]

∠ BDC = ∠ ABC   [ Both are 90° ]

∴  Δ  BDC Similar to Δ ABC   [ By AA similarity ]

So , CD / BC = BC / AC

= > BC² = CD . AC   ... ( ii )

Now adding both equation :

AB² + BC² = CD . AC +  AD . AC

AB² + BC² = AC ( CD + AD )

AB² + BC² = AC² .

AC² = AB² + BC² .

Hence proved .