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## Answers

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

The Brainliest Answer!

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