Page 21 - RMGO 3

Geometric constructions, with the compass alone, the straightedge alone or neither of them 21

from which (QE + EF) · FR = EF · (FR + RS) or QE · FR = EF · RS,
and finally
QE · FR
= RS. (3)
EF
Similarly, Intersecting Chords Theorem to AC crossed by PR in the circles
(w), (u), implies
QE · FR
= PQ. (4)
EF
From (3), (4) we have PQ = RS.
Problem 6. In the Figure 11, find (construct) the midpoint of segment AB, using
the straightedge alone.
Note: It isn’t necessary for the circles’ centers to be known to complete the
construction.

Figure 11

Solution. Let {P, Q} = (u) ∩ (v), S ∈ (v) ∩ AB and T ∈ (u) ∩ AB.

We draw segments PS, PT and QS, QT.
Then let C ∈ PS ∩ (u), D ∈ PT ∩ (v), E ∈ QS ∩ (u) and F ∈ QT ∩ (v).

We draw AE, BF and AC, BD.
Then let {K} = AE ∩ BF and {L} = AC ∩ BD.

Figure 12