proof for bpt theorem

STATEMENT = If a line is drawn parallel to one side of a triangle divides the other two sides proportionly.

  OR

  BPT

http://i25.tinypic.com/2r2olz6.jpg

GIVEN = IN TRIANGLE ABC

DE||BC

TO PROVE = AD = AE

  BD  CE

CONSTRUCTION = DRAW DM AND EN PERPENDICULAR TO AC AND AD RESPECTIVELY. JOIN BE AND CE

PROOF = ar.(TRIANGLE ADE) = 1 x AE x DM

  2

ar.(TRIANGLE DEC) = 1 x CE X DM

  2

ar.(TRIANGLE ADE) = 1 x AE x DM

ar.(TRIANGLE DEC)  2                    

  1 x CE x DM

  2

ar.(TRIANGLE ADE) = AE -------------( i )

ar.(TRIANGLE DEC)  CE

ar.(TRIANGLE ADE) = 1 x AD x EN

ar.(TRIANGLE BDE)  2 

  1 x BD x EN

  2

ar.(TRIANGLE ADE) = AD

ar.(TRINAGLE BDE)  BD

TRIANGLE BDE AND DEC ARE HAVING SAME BASE AND ARE LYING BETWEEN SAME PARALLEL'S

SO,

  ar.(TRIANGLE BDE) = ar.(TRIANGLE DEC)

ar.(TRIANGLE ADE) = AD -----------( ii )

ar.(TRIANGLE DEC)  BD

FROM ( i ) AND ( ii )

AD = AE

BD  CE

HENCE, PROVED

http://tinypic.com/r/2r2olz6/3

http://i25.tinypic.com/2r2olz6.jpg

paste it in the url box and then click on " go''  or at the green arrow button.

after pasting it you'll get the figure of this theorem

 gr8

wtf..

 tanA+cotA=2 FIND  tan⁷A+cot⁷A=

 IF tanA+cotA=2      FIND  tan⁷A+cot⁷A=

 grt

STATEMENT = If a line is drawn parallel to one side of a triangle divides the other two sides proportionly.

OR

BPT

http://i25.tinypic.com/2r2olz6.jpg

GIVEN = IN TRIANGLE ABC

DE||BC

TO PROVE = AD = AE

BD CE

CONSTRUCTION = DRAW DM AND EN PERPENDICULAR TO AC AND AD RESPECTIVELY. JOIN BE AND CE

PROOF = ar.(TRIANGLE ADE) = 1 x AE x DM

2

ar.(TRIANGLE DEC) = 1 x CE X DM

2

ar.(TRIANGLE ADE) = 1 x AE x DM

ar.(TRIANGLE DEC) 2

1 x CE x DM

2

ar.(TRIANGLE ADE) = AE -------------( i )

ar.(TRIANGLE DEC) CE

ar.(TRIANGLE ADE) = 1 x AD x EN

ar.(TRIANGLE BDE) 2

1 x BD x EN

2

ar.(TRIANGLE ADE) = AD

ar.(TRINAGLE BDE) BD

TRIANGLE BDE AND DEC ARE HAVING SAME BASE AND ARE LYING BETWEEN SAME PARALLEL 'S

SO,

ar.(TRIANGLE BDE) = ar.(TRIANGLE DEC)

ar.(TRIANGLE ADE) = AD -----------( ii )

ar.(TRIANGLE DEC) BD

FROM ( i ) AND ( ii )

AD = AE

BD CE

HENCE, PROVED

Basic Proportionality Theorem says: If a line is drawn parallel to one side of the triangle to intersect the other two sides at distinct points .Then the other two sides are divided in the same ratio. 


PROOF ( to follow this proof, just draw the triangles and segments) 
Draw triangle PQR and construct line L parallel to segment QR. 
Line L intersects segment PQ and segment PR at S and T respectively. 
We want to show that length of PS/ length of QS is equal to length PT/ length of PR since that is what the BPT says.
Construct segments SR and QT. 
Look at triangles PTS and QTS and note they have the same height which implies that 
the area of triangle PTS/ area of triangle QTS is equal to PS/ SQ. 
By the same reasoning, the areas of triangle SPT/ triangle SRT is equal to PT/TR. 
Triangles QTS and SRT both have the same height and both have ST as a base segment so they have the same area. 
So the ratio of the area of triangle PTS to the area of triangle QTS is equal to the ratios of the area of triangles SPT/SRT. 
So the ratio of PS/SQ is equal to PT/TR 

Since line L which is parallel to segment QR divides segment PQ and segment PR in the same ratio we have proved the BPT.