State & Prove Theorems
Proportionality Theorem (Thales theorem): If a line is drawn parallel to one side of a triangle intersecting other two sides, then it divides the two sides in the same ratio.
In ∆ABC , if DE  BC and intersects AB in D and AC in E then
AD AE
 = 
DB EC

Proof on Thales theorem:
If a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.
Given : In ∆ABC , DE  BC and intersects AB in D and AC in E.
Prove that : AD / DB = AE / EC
Construction: Join BC,CD and draw EF ┴ BA and DG ┴ CA.
Statements

Reasons

1) EF ┴ BA

1) Construction

2) EF is the height of ∆ADE and ∆DBE

2) Definition of perpendicular

3)Area(∆ADE) = (AD .EF)/2

3)Area = (Base .height)/2

4)Area(∆DBE) =(DB.EF)/2

4) Area = (Base .height)/2

5)(Area(∆ADE))/(Area(∆DBE)) = AD/DB

5) Divide (4) by (5)

6) (Area(∆ADE))/(Area(∆DEC)) = AE/EC

6) Same as above

7) ∆DBE ~∆DEC

7) Both the ∆s are on the same base and
between the same  lines.

8) Area(∆DBE)=area(∆DEC)

8) If the two triangles are similar their
areas are equal

9) AD/DB =AE/EC

9) From (5) and (6) and (7)

Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.
If AD AE
 =  then DE  BC
DB EC
Given: A Δ ABC and a line intersecting AB in D and AC in E,
such that AD / DB = AE / EC.
Prove that: DE  BC
Let DE is not parallel to BC. Then there must be another line that is parallel to BC.
Let DF  BC.
Statements

Reasons

1) DF  BC

1) By assumption

2) AD / DB = AF / FC

2) By Basic Proportionality theorem

3) AD / DB = AE /EC

3) Given

4) AF / FC = AE / EC

4) By transitivity (from 2 and 3)

5) (AF/FC) + 1 = (AE/EC) + 1

5) Adding 1 to both side

6) (AF + FC )/FC = (AE + EC)/EC

6) By simplifying

7) AC /FC = AC / EC

7) AC = AF + FC and AC = AE + EC

8) FC = EC

8) As the numerator are same so denominators are equal

This is possible when F and E are same. So DF is the line DE itself.
∴ DF  BC
Interior Angle Bisector Theorem :
The angle bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.
Given: A ΔABC in which AD is the internal bisector of ∠A and meets BC in D.
Prove that: BD / DC = AB / AC
Construction: Draw CE  DA to meet BA produced in E.
Statements

Reasons

1) CE  DA

1) By construction

2) ∠2 = ∠3

2) Alternate interior angles

3) ∠1 = ∠4

3) Corresponding angles

4) AD is the bisector

4) Given

5) ∠1 =∠2

5) Definition of angle bisector

6) ∠3= ∠4

6) From (2) and (3)

7) AE = AC

7) In ΔACE, side opposite to equal angles are equal

8) BD / DC = BA / AE

8) In ΔBCE DA  CE and by BPT theorem

9) BD / DC = AB / AC

9) From (7)

Pythagorean Theorem
In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
States that in a right triangle that, the square of a (a^{2}) plus the square of b (b^{2}) is equal to the square of c (c^{2}).
In short it is written as: a^{2} + b^{2} = c^{2}
Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c). Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH =
Then, we will get 4 rightangled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the
square EFGH, each of whose side is a, so area of the square EFGH is a^{2}.
Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF
or, (b + c)^{2} = a^{2} + 4 ∙ ^{1}/_{2} b ∙ c
or, b^{2} + c^{2} + = a^{2} +
or, b^{2} + c^{2} = a^{2}