if a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the divides the sides in the same proportion
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Given: In a △PQR, line l∥ side QR, line l intersect the sides PQ and PR in two distinct points M and N respectively.
To prove:
MQ
PM
=
NR
PN
... (i)
Construction: segQN and segRM are drawn.
Proof:
A(△QMN)
A(△PMN)
=
MQ
PM
(Both triangles have equal height with common vertex M)
∴
A(△RMN)
A(△PMN)
=
NR
PN
... (ii)
But A(△QMN)=A(△RMN), because they are between parallel lines MN and QR and have equal height corresponding to their common base MN ..... (iii)
From (i), (ii) and (iii), we get
A(△QMN)
A(△PMN)
=
A(△RMN)
A(△PMN)
∴
MQ
PM
=
NR
PN
[henceproved]
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