# 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.

---- = ------ 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 (a2) plus the square of b (b2) is equal to the square of c (c2).

In short it is written as: a2 + b2 = c2 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 right-angled 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 a2.

Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF

or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c

or, b2 + c2 +  = a2 +

or, b2 + c2 = a2