Question

OA and OB are respectively perpendiculars to chords CD and EF of a circle whose centre is O. If OA = OB, prove that arc EC = arc DF

Answer 1

Name the crosscut section of EF and CD as M

In figure AMBO

Angle A=90                      ------------------(1)

AO=BO                            ------------------(2)

from (1) and (2)

we get that AMBO is a square

therefore

Angle EMD=90

Angle CMF=90

therefore,

The circle is divided into 4 equal parts of 90 degree each

ar(arc ECM)=Ф\360 × 2πr         ------------(where Ф=90)

ar(arc DMF)=Ф\360 ×2πr             --------------(where Ф=90)

there,fore

Arc EC= Arc DF

Answer 2

Step-by-step explanation:

given: OA perpendicular CD

OB perpendicular EF

OA = OB

to prove:CE=DF

proof:

OA=OB

therefore CD=EF

(if the chords are equidistant from the centre then the chords are also equal)

CD=EF. ஃ arc CE = arc EF

(subtract arc ED both the side)

arc CD - arc ED =arc EF - arc ED

arc CE = arc DF

(if the corresponding arcs are equal then the corresponding chords are also equal)

ஃ CE = DF

hope it's helpful...