Answers

2015-09-13T08:28:42+05:30
Theorem; it a line is drawn parallel to one side of a triangle intersecting the other two side ,then it divides the 2 sides in the same ratio.
GIVEN
a triangle ABC in which DE//BC and intersect AB at D and AC in E
TO PROVE
AD/DB =AE/EC
CONSTRUCTION
join BE, CD and draw EFperpendicular to BA and DG perpendicular toCA
PROOF
since EF is perpendicular to AB.EF is the height of triangle ADE and DBE
now       ar(ADE)=1/2(base*hight) =1/2(AD.EF)
             ar(DBE)=1/2(DB.EF)
 there for, ar(ADE)/Ar(DBE) =1/2(AD.EF)/ar(DB.EF)=AD/DB....................1
 similarly we have
         ar(ADE)/ar(DEC)=1/2(AE.DG)/1/2(EC.DG)=AE/EC..............................2
    ar(dbe)=ar(dec)
=1/ar(DBE)=1/ar(DEc)   (by taking recipocal)
ar(ADE)/ar(DBE)= ar(ADE)/ar(DEC)
AD/DB=AE/EC               using 1 and 2
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2015-09-13T09:20:14+05:30
BASIC PROPORTION  THEOREM -IF A LINE IS PARALLEL TO ONE SIDE OF TRIANGLE TO INTERSECT OTHER TWO SIDES IN DISTINCT POINTS ,THE OTHER TWO SIDES ARE DIVIDED IN THE SAME RATIO.

TO PROVE - AD/BD=AE/CE
 
CONSTRUCTION -JOIN DC AND BE THEN DRAW DN PERPENDICULAR AC ,EN PERPENDICULAR AB.


PROVE-AREA[TRIANGLE OF ADE]  =1/2 X AD X EN     
AREA [TRIANGLE OF ADE] = 1/2 X AE X MD 

IN AREA OF BDE = 1/2 X BD X EN 
AREA OF DEC =1/2 X CE X MD 
AREA OF ADE      =1/2 X AD X EN  /1/2 X BD X EN   = AD/ BD
AREA OF BDE       

AREA OF ADE = 1/2 X AE X DM
AREA OF DEC     1/2 X CE X DM      
   
  =AE/CE

AREA OF TRIANGLE DEC=AREA OF TRIANGLE BDE 
=AD/BD=AE/CE

CONVERSE B.P.T-IF A LINE DIVIDES ANY TWO SIDES OF A TRIANGLE IN THE SAME RATIO,THEN THE LINE IS PARALLEL TO THE THIRD SIDE.

TO PROVE -DE IS PARALLEL TO THE BC.
CONSTRUCTION-DRAW ANOTHER LINE DE ' IS PARALLEL TO BC

PROOF-AD/BD =AE/EC [ GIVEN]
   ⇒  AD/BD =AE'/EC.......................(1)

AE/EC = A'E /E'C
 E AND E'  MUST  COINCIDE 
DE IS PARALLEL TO BC

⇒AE/EC +1 =AE'/E'C +1
⇒AE+EC/EC =AE'+E'C /EC

⇒EC/AC= E'C/AC
⇒EC=E'C


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