ΔABC ∼ ΔADB ∼ ΔBDC
Construction : Draw seg BD ⊥ side AC such that, A-D-C
THEOREM : Theorem of Pythagoras
In a right angled triangle, the square of the hypotenuse is
equal to the sum of the squares of the remaining two sides.
A
B
C
Given : In ΔABC, m∠ABC = 90o
To prove :
AC2
=
AC2
AB2 + BC2
AB2 + BC2
Proof :
In ΔABC,
m∠ABC = 90o
seg BD ⊥ hypotenuse AC
D
∴
ΔABC ∼ ΔADB
∴
AB
AD
=
AC
AB
= AD × AC
∴
AB2
...(i)
...(ii)
ΔABC ∼ ΔBDC
[From (i)]
[From (i)]
∴
BC
DC
=
AC
BC
= DC × AC
∴
BC2
...(iii)
Adding (ii) and (iii),
AB2 + BC2
+ (DC × AC)
= (AD × AC)
AB2 + BC2
∴
= AC
(AD + DC)
AB2 + BC2
∴
= AC
× AC
AB2 + BC2
∴
= AC2
∴
AC2
=
AB2 + BC2
[Similarity in right angled triangles]
We require squares of sides.
For that, we would require the product
And that can be obtained by using
similarity
For similarity, we require minimum 2 triangles
The existing triangle is what type of triangle ?
Right angled
triangle
That means, the triangle we require should be a right angled triangle
Let us form it
In a right angled triangle, we have perpendicular from the vertex of right angle to the hypotenuse
Which theorem can we apply?
Similarity in right
angled triangles
Let us consider 2 triangles containing side AB
Select the ratios
involving AB
Let us consider 2 triangles containing side BC
Select the ratios
involving BC
To form a right angled triangle we need to drop a perpendicular from the vertex to the opposite side
Can we drop perpendiculars from A and C ?
No
We already have it
So, we will drop a perpendicular from B to side AC
[Given]
[Construction]
[c.s.s.t]
[c.s.s.t]