Friday, May 1, 2020

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Basic Proportionality Theorem

Basic Proportionality Theorem:-

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Basic Proportionality Theorem states that "If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio".
In the following figure, segment DE is parallel to the side BC of ΔABC. Note how DE divides AB and AC in the same ratio:
Intersecting triangle proportionally

Proof of Basic Proportionality Theorem:

Given: 
  1. ΔABC
  2. DEBC
To prove: ADDB=AEEC
Construction: 
  1. Join BE and CD
  2. Draw DPAC
  3. Draw EQAB
Intersecting triangle proportionally
Proof: Consider ΔAED. If you have to calculate the area of this triangle, you can take AD to be the base, and EQ to be the altitude, so that:
ar(ΔAED)=12×AD×EQ
Now, consider ΔDEB. To calculate the area of this triangle, you can take DB to be the base, and EQ (again) to be the altitude (perpendicular from the opposite vertex E ).
Thus,
ar(ΔDEB)=12×DB×EQ
Next, consider the ratio of these two areas you have calculated:
ar(ΔAED)ar(ΔDEB)=12×AD×EQ12×DB×EQ=ADDB
In an exactly analogous manner, you can evaluate the ratio of areas of ΔAED and ΔEDC:
ar(ΔAED)ar(ΔEDC)=12×AE×DP12×EC×DP=AEEC
Finally, We know that "Two triangles on the same base and between the same parallels are equal in area". Here, ΔDEB and ΔEDC are on the same base DE and between the same parallels – DEBC .
ar(ΔDEB)=ar(ΔEDC)
Considering above results, we can note,
ar(ΔAED)ar(ΔDEB)=ar(ΔAED)ar(ΔEDC)
ADDB=AEEC
This completes our proof of the fact that DE divides AB and AC in the same ratio.

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