Let A’, B’, C‘ are contacting points of encircles to BC, AC and AB Per the result of problem 682, 3, B3 and C3 are Gergonne points AA3 will cut BC at A’ and A’B/A’C= (s-a)/(s-b) Similarly B’C/B’A=(p-a)/p-c) and C’A/C’B=(s-b)/(s-a) And A’B/A’C . B’C/B’A . C’A/C’B = 1 So AA3, BB3 and CC3 are concurrent per Ceva’s Theorem Peter Tran
Let A’, B’, C‘ are contacting points of encircles to BC, AC and AB
ReplyDeletePer the result of problem 682, 3, B3 and C3 are Gergonne points
AA3 will cut BC at A’ and A’B/A’C= (s-a)/(s-b)
Similarly B’C/B’A=(p-a)/p-c) and C’A/C’B=(s-b)/(s-a)
And A’B/A’C . B’C/B’A . C’A/C’B = 1
So AA3, BB3 and CC3 are concurrent per Ceva’s Theorem
Peter Tran
Typos:
ReplyDeleterhs's of
A’B/A’C= (s-a)/(s-b),
B’C/B’A=(p-a)/p-c)and
C’A/C’B=(s-b)/(s-a)
to be corrected as
(s-c)/(s-b),
(s-a)/(s-c)and
(s-b)/(s-a)respectively