Statement
If there are three or more parallel lines and the intercepts made by them on one transversal are equal, the corresponding intercepts of any transversal are also equal.
Given:
l||m||n
P is a transversal intersecting l,m and n at A, B and C respectively such that AB=BC.Q is another transversal drawn to cut l, m and n at D, E and F respectively. DE and EF are the intercepts made on q.
To prove:
DE = EF
Construction:
Draw a line through E parallel to p intersecting l in G, n in H respectively.
Proof:
AG||BE (given)
GE||AB (by construction)
AGEB is a parallelogram
But AB=BC (given)
From (i) and (ii), GE=EHNow compare triangles GED and EFH,
GE=EH (proved)
Hence the theorem is proved.
