Given In a circle AYDZBWCX, two chords AB and CD intersect at tight angles.
To prove arc CXA+arc DZB=arc AYD+ arc BWC =Semi-circle.
Construction Draw a diameter EF parallel to CD having centre M.
Proof Since, `CD||EF`
`:." Are"EC="arc" FD` ...(i)
Also, arc ECXA=arc EWB [ symmetrical about diameter of a circle]
and arc AF=arc BF ....(ii)
We know that, arc ECXAYDF =Semi-circle
arc EA+arc AF=Semi-circle
`rArr ` arc EC+arc CXA + arc FB=Semi-circle [from Eq. (ii)]
`rArr` arc DF+arc CXA+arc FB =Semi-circle [from Eq. (i)]
`rArr` arc DF+arc FB+ arc CXA= Semi-circle
`rArr" arc "DZB+"arc "C xx A `=Semi-circle
We know that, circle divides itself in two semi-circles, therefore the remaining portion of the circle is also equal to the semi- circle.
`:.` arc AYD+ arc BWC=Semi-circle.
