Proof 


This theorem can be proved by taking a line DE such that ${AD}/{DB}={AE}/{EC}$
and assuming that DE is not parallel to BC (see Fig.)
If DE is not parallel to BC, draw a line DE′ parallel to BC.
So, ${AD}/{DB}={AE'}/{E'C}$
Therefore ${AE}/{EC}={AE'}/{E'C}$
Adding 1 to both sides of above, you can see that E and E′ must coincide. 

Example 1: If a line intersects sides AB and AC of a ∆ABC at D and E respectively 

and is parallel to BC, prove that ${AD}/{AB}={AE}/{AC}$

Solution :


	Given: DE ||  BC     
	So, ${AD}/{DB}={AE}/{EC}$                                                              
	or, ${DB}/{AD}={EC}/{AE}$  
	Add 1 on both side, 
    ${DB}/{AD}+1={EC}/{AE}+1$ 
or	${AB}/{AD}={AC}/{AE}$

So, ${AD}/{AB}={AE}/{AC}$ ;Hence Proof

Example 2 : ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC
respectively  such  that  EF  is  parallel  to  AB,Show that ${AE}/{ED}={BF}/{FC}$               


Solution :
Join  AC  to  intersect  EF  at  G (see Fig.).



Given:AB || DC and EF || AB    
So,	EF || DC    (Lines parallel to the same line are parallel to each other)
Now, in ∆ADC,
EG ||  DC    (As EF || DC)

So, ${AE}/{ED }={AG}/{GC} ............(1)$
Similarly, from ∆CAB,
      ${CG}/{AG}={CF}/{BF}$
i.e   ${AG}/{GC}={BF}/{FC} ............(2)$
Therefore, from (1) and (2),
      ${AE}/{ED }={BF}/{FC}$

Example 3:
	In Fig.  ${PS}/{SQ}={PT}/{TR}$ and ∠PST =∠ PRQ. Prove that PQR is an isosceles triangle.
    

	
Solution :
	Given: ${PS}/{SQ}={PT}/{TR}$
	So,ST || QR                  
	Therefore, ∠ PST =  ∠ PQR    (Corresponding angles)...........(1)  
	
	Also, it is given that
	∠ PST =  ∠ PRQ      .................................(2) 
	So,  ∠ PRQ =  ∠ PQR  [From (1) and (2)]
	Therefore,  PQ =  PR         (Sides opposite the equal angles)

i.e.,    PQR is an isosceles triangle.