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 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 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 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 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