Solving Proportionality Rule

In general solving proportionality rule; categorization of proportionality for geometric ratios is a essentially serious way of say what is fundamentally obvious in the specific state of proportional ratios of numbers.  In accumulation, it may be well-known that, given incommensurable magnitudes p and q, this explanation in effect split the field of rational numbers m/n into two disjoint sets: the set L of those for which holds, or m : n < p : q, and the set U of those for which holds, or m : n > p : q.


Rule for Solving Proportionality Rule:

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio for solving proportionality rule.

Solution:

Given:

In preparation for basic proportionality theorem, a triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

Rule for Proportionality

To Prove:

(AD)/(DB) = (AE)/(EC)

Construction:

Join BE and CD and then draw DM _|_ AC and EN _|_AB.

Proof:

Area of DeltaADE = (((1)/(2)) base xx height)

So,                   area (DeltaADE) = 1/2AD xx EN

In addition,     area (DeltaBDE) = 1/2 DB xx EN.

Similarly,         area (DeltaADE) = 1/2 AE xx DM and area (DeltaDEC) = 1/2 EC xx DM.

Therefore,        (area ( Delta ADE))/(area ( Delta BDE)) = ( (1/2) AD xx EN )/( (1/2) DB xx EN ) = (AD)/(DB) rArr (i)

And                 (area ( Delta ADE))/(area ( Delta DEC)) = ( (1/2) AE xx DM )/( (1/2) EC xx DM ) = (AE)/(EC) rArr (ii)

Now the DeltaBDE and DeltaDEC is on the same base DE and between the same parallel lines BC and DE.

So,       area (DeltaBDE) = area (DeltaDEC)         rArr (iii)

Therefore, from (i), (ii) and (iii) we have,

(AD)/(DB) = (AE)/(EC).

Hence, the theorem is proved in preparation for solving proportionality rule.


Example for Solving Proportionality Rule:

In solving proportionality rule, figure, AB is parallel to CD. If OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4, determine x.

Example for Proportionality

Solution:

Given:

In a quadrilateral ABCD, AB || CD

OA = 3x – 19, OB = x – 4, OC = x – 3, OD = 4

To find:

x = ?

AB || CD

ABCD is a trapezium [The diagonals of trapezium divide each other proportionally]

(3x - 19)/(x - 3) = (x - 4)/(4)

rArr   (x – 4)(x – 3) = 4(3x – 19)

rArr   x^2 - 3x - 4x + 12 – 12x + 76 = 0

rArr   x^2 - 19x + 88 = 0                rArr   x^2 - 11x – 8x + 88 = 0

rArr   x(x – 11) – 8(x – 11) = 0         rArr   (x – 11)(x – 8) = 0

rArr   x – 11 = 0                                or                     x – 8 = 0

x = 11 units or 8 units.