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Fantezi Geometri => Fantezi Geometri => Konuyu başlatan: stuart clark - Mayıs 01, 2012, 11:57:03 öö

Başlık: orthocenter
Gönderen: stuart clark - Mayıs 01, 2012, 11:57:03 öö

If O is the orthocenter of a triangle ABC having sides BC , CA , AB

Where  BC = a , CA = b , AB = c.Then the ratio of radii of circle circumscribing the

triangle BOC , triangle COA , triangle AOB is

Başlık: Ynt: orthocenter
Gönderen: Beyşehirli - Mayıs 08, 2012, 10:51:20 öö

Let R be radii of circumcircle of ABC...
Sine Rule on ABC : BC = 2RsinA ...(1)

m(BOC) + m(A) = 180 because O is orthocenter. So sinBOC = sinA.

Let r be radii of circumcircle of OBC
Sine Rule on OBC : BC = 2rsin(BOC) = 2rsinA ...(2)

From (1) and (2) BC = 2RsinA = 2rsinA ==> R = r.

a + b + c = 2u , S(ABC) = [u(u - a)(u - b)(u - c)]^1/2 = abc/4R
==> R = r = abc / 4 [u(u - a)(u - b)(u - c)]^1/2