AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Fig. 10.26). Show that the line PQ is perpendicular bisector of AB.
Answers
Given : AB is a line segment. P and Q are points on opposite sides of AB such that AP = BP and AQ = BQ .
To prove : Show that the line PQ is perpendicular bisector of AB.
Proof: In ∆ PAQ & ∆PBQ
AP = BP (Given)
AQ = BQ (Given
PQ = PQ (Common)
Therefore, ∆ PAQ ≅ ∆PBQ (by SSS congruence rule)
From the figure ,ΔAPB and ΔABQ are two isosceles triangles in which AP = PB & AQ = BQ.
∠PAB = ∠PBA …….(1) and, ∠QAB = ∠QBA ……(2)
[Angles opposite to equal sides are equal]
In ∆ PAC & ∆ PBC
PA = PB (Given)
∠PAB = ∠PBA (from eq 1)
PC = PC (Common)
Therefore, ∆ PAC ≅ ∆PBC (by SAS congruence rule)
AC = CB …….(3) [By CPCT] and ∠PCA = ∠PCB (By c.p.c.t)
AB is a straight line :
∠PCA + ∠PCB = 180°
∠PCA + ∠PCA = 180°
[∠PCA = ∠PCB]
2∠PCA = 180°
∠PCA = 180°/2
∠PCA = 90° ………(4)
From eq 3 & 4 we can say that C is the point on line PQ, and PQ is the perpendicular bisector of AB.
Hence, PQ is the perpendicular bisector of AB.
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AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A...
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