1.

If two chords AB and CD of a circle AYDZBWCX intersect at right angles,then prove that arc CXA+arc DZB=arc AYD+arc BWC =semi-circle.

Answer»

Solution :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,arcECXA=arcEWB[ SYMMETRICAL about diameter of a circle]
andarc 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.


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