ABCD is a rhombus. CA is the circle through B, C, D; CB is the circle through A, C, D; CC is the circle through A, B, D; and CD is the circle through A, B, C. Show that the angle between the tangents to CA and CC at B equals the angle between the tangents to CB and CD at A.
Solution
The fact that ABCD is a rhombus is irrelevant. The result holds for any quadrilateral.
We can mark the angles as shown. We wish to show that c = e. We have:
a + b + c = 180o (looking at A) (1)
a + d + e = 180o (looking at B) (2)
c + d + f = 180o (looking at C) (3)
b + f + e = 180o (looking at D) (4)
Taking (1) - (2), we get c + e + b - d = 0 and taking (3) - (4) we get c + e + d - b = 0. Adding gives c = e as required.
Note that the derivation still holds for different configurations, such as
© John Scholes
jscholes@kalva.demon.co.uk
29 Oct 2003
Last corrected/updated 29 Oct 03