if a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the sides in the same proportion. Prove it
Answers
Given: The theorem: if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points , then the other two sides are divided in the same ratio.
To find: Prove the theorem (Basic Proportionality Theorem)
Solution:
- First we need to do some constructions.
- In triangle PQR, let l be a line parallel to QR, let the intersected points by l be M on PQ and N on PR, and now join QN and MR.
- Now After looking to the figure, we get:
area( tri MPN ) / area ( tri NQM ) = PM / MQ ..............(i)
- As both triangles have the height same (MN) and a common vertex (M).
- Similarly :
area( tri MPN ) / area ( tri NRM ) = PN / NR .................(ii)
- Now:
area( tri NQM ) = area ( tri NRM ) .........(iii)
(triangle lies in parallel lines and have same base)
- Now from equation i, ii and iii, we have:
area( tri MPN )/ area ( tri NQM ) = area( tri MPN ) = area ( tri NRM )
- So similarly:
PM / MQ = PN / NR
Hence proved.
Answer:
So in the above solution we proved the theorem of Basic Proportionality Theorem Or Thales Theorem.
A line is parallel to one side of triangle which intersects remaining two
sides in two distinct point then that line divides the sides in same proportion.
Given: In ∆ABC line l II side BC & line l intersect side AB in P & side AC in Q
To Prove:
=
, Construction: Draw CP & BQ.
Proof: ∆APQ&∆PQB have equal heights
(∆)
(∆)
=
(areas proportionate to bases) ----I
(∆)
(∆)
=
(areas proportionate to bases) -- II
Seg is common base of ∆PQB & ∆PQC. SegPQ II Seg BC hence ∆PQB &
∆PQC have equal heights