SECTION 1.2 Triangle Geometry 7
- In the figure to the right,ABCDis
a parallelogram, and E is a point
on the segment [AD]. The point
F is the intersection of lines (BE)
and (CD). Prove thatAB×FB=
CF×BE. - In the figure to the right, tangents
to the circle atBandCmeet at the
point A. A point P is located on
the minor arcBC ̆ and the tangent
to the circle at P meets the lines
(AB) and (AC) at the pointsDand
E, respectively. Prove thatDOÊ =
1
2 B
OĈ , whereOis the center of the
given circle.
1.2.4 “Sensed” magnitudes; The Ceva and Menelaus theo-
rems
In this subsection it will be convenient to consider the magnitudeABof
the line segment [AB] as “sensed,”^2 meaning that we shall regardAB
as being either positive or negative and having absolute value equal to
the usual magnitude of the line segment [AB]. The only requirement
that we place on the signed magnitudes is that if the pointsA, B,and
Care colinear, then
AB×BC=
> 0 if−→
AB and−→
BC are in the same direction
< 0 if−→
AB and−→
BC are in opposite directions.(^2) IB uses the language “sensed” rather than the more customary “signed.”