Prove that in a right angled triangle the square of hypotenuse is equal to the sum of the squares of the
other two sides
Answers
Answer:
In the diagram above,
∠ABC is a right angle.
AC is the hypotenuse.
AB is known as the perpendicular.
BC is the base.
So according to the Pythagoras Theorem,
(AC)²=(AB)²+(BC)²
But then should we merely trust a single statement? I don’t think so. We need a proof!
Proof of the Pythagoras Theorem using Similarily Given: In ΔABC, m∠ABC=90°
Construction: BD is a perpendicular on side AC
To Prove: (AC)²=(AB)²+(BC)²
Proof:
In △ABC,
m∠ABC=90° (Given)
seg BD is perpendicular to hypotenuse AC (Construction)
Therefore, △ADB∼△ABC∼△BDC
(Similarity of right-angled triangle)
△ABC∼△ADB
(AB/AC)=(AD/AB) (congruent sides of similar triangles)
AB2=AD×AC (1)
△BDC∼△ABC
CD/BC=BC/AC (congruent sides of similar triangles)BC2=CD×AC (2)
Adding the equations (1) and (2),
AB2+BC2=AD×AC+CD×AC
AB2+BC2=AC(AD+CD)
Since, AD + CD = AC
Therefore, AC2=AB2+BC2
Hence Proved.
There also exists a Converse of the Pythagoras theorem that states, “If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle”.
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 .
