If two chords AB and CD of a circle AYDZBWCX intetsect at right angles (see fig.), prove that arc CXA+arc DZB = arc AYD + arc BWC= semicircle
please be fast!!
I urgently need the answer
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deepak70:
ok.
yes ncert examlpler is a different book
hmm.
Its of which class.......9th or 10th
this question is frm 9th btw book dono hai 9th bhi 10th bhi
i know the answer but dont know the way to explain it
oh i m in 9 th so i donot know the answer.
main bhi 9th mein hoon par yr i dont know to explain it
to tumhe iska jo 10 th ka concept h wo kaise pt
pta*?
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Answered by
47
Hope it works..........
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Answered by
32
draw a diameter EF ll to CD
since CD ll EF
therefore arc EC=arc FD .....(1)
also, arc ECXA=arc EWB .....diameter as line of symmetry
and,arc AF= arc FB .....(2)
since arc ECXAYDF = semicircle...
therefore arc EA + arc AF = semicircle
=> arc EC +arc CXA +arc FB= semicircle....by (2)
=>arc DF + arc FB +arc CXA= semicircle....by (1)
=> arc DZB +arc CXA= semicircle
since a circle divides itself in two semicircles
therefore the remaining part , arc AYD +arc BWC=semicircle
since CD ll EF
therefore arc EC=arc FD .....(1)
also, arc ECXA=arc EWB .....diameter as line of symmetry
and,arc AF= arc FB .....(2)
since arc ECXAYDF = semicircle...
therefore arc EA + arc AF = semicircle
=> arc EC +arc CXA +arc FB= semicircle....by (2)
=>arc DF + arc FB +arc CXA= semicircle....by (1)
=> arc DZB +arc CXA= semicircle
since a circle divides itself in two semicircles
therefore the remaining part , arc AYD +arc BWC=semicircle
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Thanku so much
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