If a line is drawn parallel to one side of a triangle to intersect the other teo sides in distinct points , the other 2 sides are divided in the same ratio
Answers
Basic Proportionality Theorem
In the figure alongside, if we consider DE is parallel to BC, then according to the theorem,
ADBD=AECE
Let’s not stop at the statement, we need to find a proof that its true. So shall we begin?
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.
