Question

state and prove the Basic proportionality theorem

Answer 1

Thales theoram of triangle
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

Answer 2

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