Question

In the given figure, P is the midpoint of arc ABP and M is the midpoint of chord AB of a circle with centre as O. Prove that: (i) PM is perpendicular to AB (ii) PM produced will pass through the centre O
(iii) PM produced will bisect the major arc AB.

Answer 1

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.

Answer 2

Answer: Triangle is construction.

Step-by-step explanation:

The fundamental language that connects everything is construction math. You must master the fundamental language if you wish to improve your abilities. Measurements are a crucial component of construction math.

Given

A circle with a center '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

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    Linear 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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