Pythagoras and Thales Theorem explained

1. PYTHAGORAS
THEOREM
ANURAG
X-B
R.NO-54

2. Pythagoras of Samos
 Pythagoras (569-479 B.C) was an Ionian
Greek philosopher.
 He was born on the Island of Samos.
 He travelled Egypt, Greece and India &
returned to Samos in 530 B.C.
 He made influencial contribution to
philosophy and religion in late 6th century.
 He is often revered as a great
mathematician and scientist ; best known
for the Pythagorean theorem.

3. The Pythagoras Theorem
 In a right angled triangle, the square of
the hypotenuse is equal to the sum of the
squares of the other two sides
BASE
P
E
R
P
E
N
D
I
C
U
L
A
R
A B
C
BC2 = AC2 +
AB2

4. PROOF:
A C
B
D
•We are given a right
angled triangle ABC
right angled at B.
•We need to proved
that AC2 = BC2 + AB2
• Let us draw BD ┴ AC
• Now ▲ADB~ ▲ABC
(as per theorem 6.7)
• So, AD = AB(sides prop)
AB AC

5. PROOF:
A C
B
D
• Or, AD X AC = AB2 …………………… (1)
• Also, ▲BDC~ ▲ABC (Theorem 6.7)
• So, CD = BC(sides prop)
BC AC
• CD X AC = BC2 …………………… (2)
• Now, let us add both Equation (1) & (2)
• ADXAC + CDXAC = AB2 + BC2
• AC(AD+CD) = AB2 + BC2
• AC2= AB2 + BC2

6. Thank You !

7. THALES
THEOREM
ANURAG
X-B

8.  Thales(624-546 B.C) was a pre-socratic
Greek philosopher and mathematician
from Miletus in Asia Minor & one of the
seven sages of Greece.
 In Mathematics, Thales used geometry to
calculate the heights of pyramids and
distance of ships from the shore.

9. The Thales Theorem
 If a line is drawn parallel to the third side
of a triangle to intersect the other two
sides in distinct points, the other two
sides are divided into the same ratio.
 i.e. in ▲ABC, side DE ǁ BC,
• Then, AD= AE
DB EC
A
CB
D E

10. PROOF:
 In ▲ABC, side DE ǁ BC, we need to prove that
AD= AE
DB EC
 Let us join BE and CD, then draw
DM ┴ AC & EN ┴ AB
 Area of ▲ADE= 1 ADXEN
2
 i.e. Ar(ADE)= 1 ADXEN
2
 Ar(BDE)= 1 DBXEN
2
A
CB
D E
N M

11.  Ar(ADE)= 1 AEXDM
2
 Ar(DEC)= 1 ECXDM
2
 Ar(ADE) = ½ AD X EN = AD ……(1)
Ar(BDE) ½ DB X EN DB
 Ar(ADE) = ½ AE X DM = AE ……(2)
Ar(DEC) ½ EC X DM EC
 ▲BDE & ▲ DEC are on the same base DE &
between the same parallel BC and DE,
 So, Ar (BDE) = Ar (DEC) …….(3)
• From (1), (2) & (3) AD= AE
DB EC
A
CB
D
E
N
M

12. PPREPARED BY ANURAG AND
MOHIT
THANK YOU !