### Dictionary Rank of a Word | Permutations & Combinations

PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

### Basic Proportionality Theorem (BPT), Thales Theorem

Basic Proportionality Theorem Class 10th
(OR)  B.P.T.   or Thales Theorem
Converse of basic proportionality theorem, thales theorem 10th standard, theorem 6.2 class 10

Statement:-
If a line is drawn parallel to one side of the triangle to intersect the other two sides in two distinct points, the other two sides are divided in the same ratio.
Given:-
A  Δ ABC in which line  BC, intersect side AB and AC at point D and E

To Prove :- $\frac{AD}{DB}=\frac{AE}{EC}$
Construction :-
Draw  EM AB and DN AC. Also join BE and CD
Proof :- $Area\: \: \: of\: \: triangle = \frac{1}{2}\times Base\times Height$
$Area\: \: \: of\: \:\Delta ADE=\frac{1}{2}\times AD\times EM............(1)$
$Area\: \: \: of\: \:\Delta BDE=\frac{1}{2}\times BD\times EM .......... (2)$
$Dividing\: \: equation \: \: (1)\: \: by\: \: equation\: \: (2)$
$\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{\frac{1}{2}\times AD\times EM}{\frac{1}{2}\times BD\times EM}$

$\frac{Ar(\Delta ADE)}{Ar(\Delta BDE)}=\frac{AD}{BD}\: \: \: ....................(3)$

Similarly
$\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{\frac{1}{2}\times AE\times DN}{\frac{1}{2}\times EC\times DN}$
$\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}=\frac{AE}{CE}\: \: \: \: ....................(4)$

ΔBED and ΔCDE have same base and lie between the same parallel DE and BC
Ar(ΔBED) = Ar(ΔCDE)   ...........................................(5)

From equation  (3), (4), (5) ...........
$\frac{Ar(\Delta ADE)}{Ar(\Delta BED)}=\frac{Ar(\Delta ADE)}{Ar(\Delta CDE)}$
$\frac{AD}{DB}=\frac{AE}{EC}$

Hence prove the Basic Proportionality Theorem.

Note:- For the examination point of view students should study the basic proportionality theorem only its converse is only a motivational theorem.

Converse of Basic Proportionality Theorem

Statement:-

If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.

Given:-  $In\: \Delta ABC,\:\frac{AD}{DB}=\frac{AE}{EC}$

To Prove :-  Line DE BC

Construction:-

If DE is not parallel to BC, then let us take another line DE' BC
Proof:-

In ΔABC,  DE' BC Therefore by B.P.T
$\frac{AD}{DB}=\frac{AE'}{E'C}$
$But\: \frac{AD}{DB}=\frac{AE}{EC}\: \: ............Given$
$Therefore\: \: \: \: \: \: \frac{AE}{EC}=\frac{AE'}{E'C}$
$1+\frac{AE}{EC}=1+\frac{AE'}{E'C}$
$\frac{EC+AE}{EC}=\frac{E'C+AE'}{E'C}$
$\frac{AC}{EC}=\frac{AC}{E'C}$
$EC =E'C$
This is possible only if E and E' coincide with each other

⇒ E and E' represent the same point on the side of the triangle.

Hence DE is parallel to the side BC

Converse of BPT is proved

Important result based on BPT

If a line intersects side AB and AC of a ΔABC at D and E respectively and is parallel to BC, then prove that $\frac{AD}{DB}=\frac{AE}{AC}$

Solution :-
It is given that DE॥BC, therefore by BPT$\frac{AD}{DB}=\frac{AE}{EC}$
$\frac{DB}{AD}=\frac{EC}{AE}\: \: \: .........By\: \: Invertendo$
Adding 1 on both side we get
$\frac{DB}{AD}+1=\frac{EC}{AE}+1$
$\frac{DB+AD}{AD}=\frac{EC+AE}{AE}$
$\frac{AB}{AD}=\frac{AC}{AE}$
$\frac{AD}{AB}=\frac{AE}{AC}\: \: \: \: ............By\: \: Invertendo$