# Prove basic proportionality theorem class 10 and converse of them also..

2
by sahilverma

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
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
ar(DBE)=1/2(DB.EF)
similarly we have
ar(dbe)=ar(dec)
=1/ar(DBE)=1/ar(DEc)   (by taking recipocal)
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.

CONSTRUCTION -JOIN DC AND BE THEN DRAW DN PERPENDICULAR AC ,EN PERPENDICULAR AB.

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 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

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

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

PLZ MARK BRAINLEST.
ok Nice it belongs yo you