5.2: Perpendicular Bisector
( \newcommand{\kernel}{\mathrm{null}\,}\)
Assume M is the midpoint of the segment [AB]; that is, M∈(AB) and AM=MB.
The line ℓ that passes thru M and perpendicular to (AB), is called the perpendicular bisector to the segment [AB].
Given distinct points A and B, all points equidistant from A and B and no others lie on the perpendicular bisector to [AB].
- Proof
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Let M be the midpoint of [AB].
Assume PA=PB and P≠M. According to SSS (Theorem 4.4.1), △AMP≅△BMP. Hence
∡AMP=±∡BMP.
Since A≠B, we have "-" in the above formula. Further,
π=∡AMB≡≡∡AMP+∡PMB≡≡2⋅∡AMP.
That is, ∡AMP=±π2. Therefore, P lies on the perpendicular bisector.
To prove the converse, suppose P is any point on the perpendicular bisector to [AB] and P≠M. Then ∡AMP=±π2, ∡BMP=±π2 and AM=BM. By SAS, △AMP≅△BMP; in particular, AP=BP.
Let ℓ be the perpendicular bisector to the segment [AB] and X be an arbitrary point on the plane.
Show that AX<BX if and only if X and A lie on the same side from ℓ.
- Hint
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Assume X and A lie on the same side of ℓ.
Note that A and B lie on opposite side of ℓ. Therefore, by Corollary 3.4.1, [AX] does not intersect ℓ and [BX] intersects ℓ; suppose that Y denotes the intersection point.
Note that BX=AY+YX≥AX. Since X∉ℓ, by Theorem 5.2.1 we have BX≠BA. Therefore BX>AX.
This way we proved the "if" part. To prove the "only if" part, you need to switch A and B and repeat the above argument.
Let ABC be a nondegenerate triangle. Show that
AC>BC⇔|∡ABC|>|∡CAB|.
- Hint
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Apply Exercise 5.2.1, Theorem 4.2.1 and Exercise 3.1.2.