Transcript Slide 1

Theorem 11. If 3 parallel lines cut off equal segments on some transversal
line, then they cut off equal segments on any other transversal.
USE THE FORWARD AND THE BACK ARROWS ON THE KEYBOARD TO VIEW AND REWIND PROOF.
Given: 3 parallel lines f, g and h intersecting a transversal
line k at points A, B and C such that |CB|= |BA|.
Another transversal line t intersects the parallel lines at D,
E and F.
To Prove: |DE|= |EF |
Construction: (i)Draw AM parallel to DF and
(ii)Draw MP parallel to CA
Proof: |MQ| = |CB| Opposites sides of a parallelogram
But |CB| = |BA| Given
Therefore |MQ| = |BA|
|BAN| = |NMQ| Alternate angles between parallel lines
|BNA| = |MNQ| Vertically Opposite Angles
Therefore |ABN| = |NQM|
Triangle ABN is congruent to triangle MNQ (ASA)
Therefore |AN| = |NM|
But |AN| = |DE| and |NM|= |EF| Opposite sides of parallelograms
Therefore |DE|= |EF | Q.E.D.
© Project Maths Development Team