Download Lesson 125: Distance Defined, Equidistant from Two Points

Lesson 125:
Distance Defined, Equidistant
from Two Points, Equidistant
from Sides of an Angle, Circle
Proofs
Distance Defined:
We remember the basic
definition of distance:
The distance between two
points is the length of the
segment that connects the
points.
The distance from a point
to a line is measured
along the perpendicular
between the point and the
line.
Equidistant from two points:
Any two points that are equidistant
from the ends of a segment lie on the
perpendicular bisector of the
segment. We will prove this in two
ways: once in paragraph form and
once in column form.
On the left we show segment BD
and two points, A and C, that are
equidistant from B and D. we use
these points to draw the figure on
the right.
The double tick marks indicate equal
distances from A, and the single tick
marks indicate equal distances from
C. Segment AC is a side of both
ΔABC and ΔADC. Thus these
triangles are congruent by SSS. This
means that the two angels at A are
equal by CPCTC. We use tick marks
to mark these angles as equal angles
and focus on the left hand part of the
figure.
The upper and lower triangles are congruent
by SAS, because AE is a side of both
triangles. Thus BE ≅ ED by CPCTC. Also, the
two angles at E are equal by CPCTC, which
means they are right angles (because they
are equal and form a straight angle). This
means that AE is perpendicular to BD and
bisects BD.
Now we will draw a two column
proof based on the figures and
reasoning used before.
Equidistant from sides of an angle:
Any point that is equidistant from
the two sides of an angle lies on
the ray that is the angle bisector.
On the left we show point X that is
equidistant from the sides of the
angle BAC. On the right we draw
perpendicular segments whose
lengths equal these distances and
also draw ray AX.
Segment AX is the hypotenuse of
both right angles and is congruent to
itself. Thus the two triangles are
congruent by HL, and angle BAX is
congruent to angle CAX by CPCTC.
Since these two angles have equal
measures and are adjacent angles,
ray AX is the bisector of angle BAC,
and X lies on the angle bisector.
This proof can be repeated for any
point that is equidistant from the
sides of an angle.
We see that the locus of points
that are equidistant from the sides
of an angle is the ray that bisects
the angle.
Circle proofs:
Proofs involving the relationship between
circles, chords, and tangents often use the
fact that all radii of a circle have equal
lengths. If two circles have radii whose
lengths are equal, the circles are congruent
circles. This statement is a definition, so we
can also say that if two circles are
congruent, their radii have equal lengths.
Another useful fact is that a
tangent to a circle is perpendicular
to a radius of the circle at the point
of tangency.
This fact can be proved, but the
methods required are beyond the
scope of this book. However, we
present many other theorems about
segments and circles and prove them
by showing that two triangles are
congruent.
The lengths of tangent segments that
intersect at a point outside the circle are
equal. In the left hand figure we show
two tangents to a circle O that intersect
at point P. in the right hand figure we
draw radii to the points of tangency and
draw segment OP.
The radii are the legs of the two
right triangles. Segment OP is the
hypotenuse of both right triangles.
Thus the triangles are congruent
by HL, and AP ≅ BP by CPCTC.
If a radius is perpendicular to a chord,
then it bisects the chord. The figure on
the left shows a radius that is
perpendicular to a chord. In the right
hand figure we draw radii to A and B.
Since OE is a leg of both right
triangles, the triangles are congruent
by HL. Thus the two angles at E are
congruent and AE ≅ EB by CPCTC.
Because the angles are equal and
sum to 180°, they are right angles.
Thus OD and OE are perpendicular
bisectors of AB.
If a radius bisects a chord that is not
a diameter, it is perpendicular to that
chord. On the left we show the basic
figure. In the right hand figure we
draw radii to E and F.
Segment OG is congruent to itself.
Thus these triangles are congruent
by SSS. It follows that angles EGO
and FGO are congruent by
CPCTC. Because they sum to
180°, these angles are also right
angles.
The perpendicular bisector of a
chord passes through the center of
the circle. Again we draw two radii
in the right hand figure.
This time we do not have to prove
that two triangles are congruent. All
we have to do is remember that we
have already proved that two points
that are equidistant from the sides of
a segment lie on the perpendicular
bisector of the segment. Since O is
equidistant from C and D, it lies on
the perpendicular bisector of CD.
HW: Lesson 125 #1-30