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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. |
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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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