# Question) In the given figure, P is the mid-point of arc APB and M is the mid-point of chord AB of a circle with centre O. . Prove that : (i) $PM\perp AB\phantom{\rule{0ex}{0ex}}$ (ii) PM produced will pass through the centre O; (iii) PM produced will bisect the major arc AB.

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Given
A circle with centre 'O' in which AB is a chord lying in the circle.
'P' is the mid-point of minor arc AB and 'M' is the mid-point of chord AB.

To Prove
1. PM is perpendicular to AB

Construction
Construct two chords by joining AP & PB

​Proof
In Triangle AMP & Triangle BMP
MP=MP        (Common Side)
AP=BP          (Since, 'P' is the mid-point of minor arc AB)
AM=BM       (Since, 'M' Is the mid-point of chord AB)
By S-S-S Congruence rule
Triangle AMP is congruent to Triangle BMP
Therefore, Angle AMP= Angle BMP  (C-P-C-T)
Since,
Angle AMP+ Angle BMP= 180    (Lenear pair)
2Ang AMP= 180     (Since, AMP=BMP, Proved Above)
Angle AMP= 180/2
AMP= 90
Since, Both the angles are 90 degrees therefore, PM is the perpendicular bisector of AB.

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