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AB'= C'A(each = s)B'C = CA'(each = s - b)C'B = BA'(each = s - c)Follows(AB'/B'C).(CA'/A'B).(BC'/C'A)=(AB'/C'A).(CA'/B'C).(BC'/A'B)= 1.1.1= 1Hence by Ceva's TheoremAA', BB', CC'are concurrent
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AB'= C'A(each = s)
ReplyDeleteB'C = CA'(each = s - b)
C'B = BA'(each = s - c)
Follows
(AB'/B'C).(CA'/A'B).(BC'/C'A)
=(AB'/C'A).(CA'/B'C).(BC'/A'B)
= 1.1.1
= 1
Hence by Ceva's Theorem
AA', BB', CC'are concurrent