If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.
Solution:
Given, the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q.
We have to prove that arc PXA ≅ Arc PYB.
PQ is the perpendicular bisector of AB
So, AM = B M
Considering triangle APM and BPM,
Since PQ is the perpendicular bisector, M is the midpoint of AB
AM = BM
∠AMP = ∠BMP = 90°
Common side = MP
SAS criterion states that if two sides of one triangle are proportional to two sides of another triangle and their included angles are congruent, then the triangles are similar.
By SAS criterion, the triangles APM and BPM are similar.
The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are similar, then their corresponding sides and angles are also congruent or equal in measurements.
By CPCTC,
AP = BP
We know that if two chords of a circle are congruent, then their corresponding arcs are equal.
Therefore, arc PXA = arc PYB
✦ Try This: The radius of a certain circle is 30 cm. Find the approximate length of an arc of this circle; if the length of the chord of the arc be 30 cm.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 2
If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.
Summary:
If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, it is proven that arc PXA ≅ Arc PYB
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