Perpendicular Bisector Proof

Introduction to perpendicular bisector proof:


Aperpendicular segment is defined as the segment which passes throughoutthe medium that is referred as the perpendicular bisector. The Separation of a line into two equal segments is made by perpendicular bisector. The perpendicular bisector proof are also applicable for the several forms like locus of the lines, circumcentres of the cicles, chords, etc.



Perpendicular Bisector Proof:


Theoretical proof:

In WXY, let D divide XY in the ratio WX/WY. We have to prove that WD bisects ∠XWY .


Proof:

Letus move to prove the statements which are given to satisfy the perpendicular bisector proof.. Since WD parallel to YE, XD/DY = XW/WE. But since XD/DY = XW/WY, WE="WY." Hence ∠WEY =∠WYE. Since ∠WEY =∠XWD (corresponding angles) and ∠WYE =∠DWY (alternate angles), it follows that ∠XWD = ∠DWY.

By the same way, if D separates XY externally in the ratio WX/WY, after we have that WD is the external bisector of ∠XWY.

Perpendicular Bisectors Proof for the Applications:


For locus of a point:


The locus of a point equidistant from two fixed points is the perpendicular bisector of the line segment joining the two points.


Proof:



Consider the lines l1and l2 as the paperFold the paper such that the lines fall on each other. Every point on the fold is equidistant from the two lines. Hence we may say the locus of a point equidistant from two given parallel lines is the parallel line situated midway between the lines.


For the sides of a triangle:


Generallythe perpendicular bisector of a side of a triangle is the line perpendicular to the side and it also bisects it. A triangle has only three sides, and considering for each side, there is a perpendicular bisectors of its sides. In the figure there is a ΔABC. Line one is the perpendicular bisector of BC.


Solution:


Drawthe triangle Δ ABC. Draw the perpendicular bisects of sides BC and CA. Plot the point of intersection of the two perpendicular bisectors as O. Draw OR-AB to rule AB in R. Measure AR and RB.



Wewould observe that AR = RB, so that OR is the perpendicular bisector ofAB. Thus, the point O is common to three perpendicular bisectors of thesides of ∠ABC. Repeat the above procedure with two others Δs We would observe that, in each case, the perpendicular bisector of the sides of the pass through a common point.


Conclusion:


The perpendicular bisectors for the sides of the triangle is called the circum-centre of the triangle. Thus; for to locate the circum-centre of atriangle it is sufficient to draw perpendicular bisectors of the two sides only.