Math, asked by bebita, 1 year ago

verification of basic proportionality theorem.

Answers

Answered by sriram007
12
PROOF OF BPT

Given: In  ΔABC, DE is parallel to BC

Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.

To Prove: ADBD=AECE

Construction: Join segments DC and BE

Proof: 

In ΔADE and ΔBDE,

A(ΔADE)A(ΔBDE)=ADBD                 (triangles with equal heights)

In ΔADE and ΔCDE,

A(ΔADE)A(ΔCDE)=AECE                  (triangles with equal heights)

Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,

A(ΔBDE)=A(ΔCDE)

Therefore,

A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)

Therefore,

ADBD=AECE

Hence Proved.

Answered by deepshika1512
5
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Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. ... Basic Proportionality Theorem (can be abbreviated as BPT) states that, if a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion

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bebita: thnk u but i want the verify not defination....☺☺☺
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