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) P M A B
(ii) PM produced will pass through the centre O;




(iii) PM produced will bisect the major arc AB.

Dear student
We have,arcAP=arc PBAP=PB   [Chords of congruent arcs are equal]OP is the bisector of APBOPA=OPBThus, inAMP and BMP, we have,PM=PM [Common]AP=PBOPA=OPBSo, by SAS congruence criterion, we haveAMPBMPAM=BM and AMP=BMPBut AMP+BMP=180°AMP=BMP=90°Hence PM or PO is the prependicular bisector of 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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