When a heat divides one more line segment into two same halves through its midpoint at 90º, that is called the perpendicular of the line segment. The perpendicular bisector theorem states that any suggest on the perpendicular bisector is equidistant native both the endpoints the the line segment ~ above which that is drawn. If a column is standing at the center of a bridge at one angle, all the point out on the column will it is in equidistant native the end points of the bridge.
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|1.||What is a Perpendicular Bisector?|
|2.||What is Perpendicular Bisector Theorem?|
|3.||What is the Converse that Perpendicular Bisector Theorem?|
|4.||Proof that Perpendicular Bisector Theorem|
|5.||Solved examples on Perpendicular Bisector Theorem|
|6.||Practice questions on Perpendicular Bisector Theoerem|
|7.||FAQs top top Perpendicular Bisector Theorem|
What is a Perpendicular Bisector?
A perpendicular bisector is a heat segment the intersects another line segment in ~ a appropriate angle and also it divides that various other line right into two equal components at that midpoint.
What is Perpendicular Bisector Theorem?
The perpendicular bisector theorem claims that any suggest on the perpendicular bisector is equidistant from both the endpoints of the line segment ~ above which the is drawn.
In the above figure,
MT = NT
MS = NS
MR = NR
MQ = NQ
What is the Converse of Perpendicular Bisector Theorem?
The converse that the perpendicular bisector theorem states that if a allude is equidistant native both the endpoints the the line segment in the exact same plane, then that allude is top top the perpendicular bisector of the line segment.
In the above image, XZ=YZ
It implies ZO is the perpendicular bisector the the line segment XY.
Proof of Perpendicular Bisector Theorem
Let us look at the proof of the over two theorems on a perpendicular bisector.
Perpendicular Bisector theorem Proof
Consider the complying with figure, in which C is one arbitrary allude on the perpendicular bisector of AB (which intersects AB at D):
Compare \(\Delta ACD\) and \(\Delta BCD\). We have:AD = BDCD = CD (common)∠ADC =∠BDC = 90°
We watch that \(\Delta ACD \cong \Delta BCD\) by the SAS congruence criterion. CA = CB,which way that C is equidistant native A and also B.
Note: refer to the SAS congruence criterion to know why \(\Delta ACD\) and \(\Delta BCD\) are congruent.
Perpendicular Bisector organize Converse Proof
Consider CA = CB in the over figure.
To prove that ad = BD.
Draw a perpendicular line from suggest C the intersects heat segment abdominal at allude D.
Now, compare \(\Delta ACD\) and \(\Delta BCD\). We have:AC= BCCD = CD(common)∠ADC = ∠BDC = 90°
We view that \(\Delta ACD \cong \Delta BCD\) by the SAS congruence criterion. Thus, ad = BD, which means that C is equidistant native A and B.
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Important NotesThe perpendicular bisector theorem and its converse deserve to be showed by the SAS congruency criterion.The perpendicular bisector theorem is offered in the construction of buildings, bridges, etc., and in making designs where we require to construct something in the center and at equal street from the endpoints.
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