# BASIC PROPORTIONALITY THEOREM PROOF

Theorem :

If a straight line is drawn parallel to one side of a triangle intersecting the other two sides, then  it divides the two sides in the same ratio.

Given : In a triangle ABC, a straight line parallel to BC, intersects AB at D and AC at E.

To prove : AD/DB = AE/EC

Construction :

Join BE, CD.

Draw EF ⊥ AB and DG ⊥ CA ## Proof :

Step 1 :

Because EF ⊥ AB, EF is the height of the triangles ADE and DBE.

Area (ΔADE) = 1/2 ⋅ base ⋅ height  =  1/2 ⋅ AD ⋅ EF

Area (ΔDBE) = 1/2 ⋅ base ⋅ height  =  1/2 ⋅ DB ⋅ EF

Therefore,

Area (ΔADE) / Area (ΔDBE) :

= (1/2 ⋅ AD ⋅ EF)/(1/2 ⋅ DB ⋅ EF)

Step 2 :

Similarly, we get

Area (ΔADE) / Area (ΔDCE) :

= (1/2 ⋅ AE ⋅ DG)/(1/2 ⋅ EC ⋅ DG)

Area (ΔADE)/Area (ΔDCE) = AE/EC ----(2)

Step 3 :

But ΔDBE and ΔDCE are on the same base DE and between the same parallel straight lines BC and DE.

Therefore,

Area (ΔDBE) = Area (ΔDCE) ----(3)

Step 4 :

From (1), (2) and (3), we can obtain

Hence, the theorem is proved.

Examples 1-2 : Find the missing length indicated.

Example 1 : Solution :

In the triangle shown above, line segment ED is parallel to the side AC.

By Triangle Proportionality Theorem,

BE/EA = BD/DC

Substitute.

15/EA = 10/4

15/EA = 5/2

Take reciprocal on both sides.

EA/15 = 2/5

Multiply both sides by 15.

EA = 6

Example 2 : Solution :

In the triangle shown above, line segment ED is parallel to the side AB.

By Triangle Proportionality Theorem,

CD/DA = CE/EB

CD/DA = (CB - EB)/EB

Substitute.

15/DA = (12 - 2)/2

15/DA = 10/2

15/DA = 5

Take reciprocal on both sides.

DA/15 = 1/5

Multiply both sides by 15.

DA = 3

Example 3 :

Solve for x. Solution :

In the triangle shown above, line segment ED is parallel to the side BC.

By Triangle Proportionality Theorem,

(AB - DB)/DB = AE/EC

Substitute.

(28 - 8)/8 = (3x - 5)/10

20/8 = (3x - 5)/10

5/2 = (3x - 5)/10

Multiply both sides by 10.

25 = 3x - 5

30 = 3x

Divide both sides by 3.

10 = x

Example 4 :

Solve for x. Solution :

In the diagram above, line segments AD, BE and CF are parallel to each other.

By Theorem,

AB/BC = DE/EF

Substitute.

15/6 = 25/EF

5/2 = 25/EF

Take reciprocal on both sides.

2/5 = EF/25

Multiply both sides by 25.

10 = EF

Example 5 :

Solve for x. Solution :

In the diagram above, line segments AD, BE and CF are parallel to each other.

By Theorem,

AB/BC = DE/EF

AB/BC = DE/(DF - DE)

Substitute.

(2x - 10)/9 = 4/(10 - 4)

(2x - 10)/9 = 4/6

(2x - 10)/9 = 2/3

3(2x - 10) = 9(2)

6x - 30 = 18

6x = 48

Divide both sides by 6.

x = 8

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