Crazy Geometry (3)

With ref. to the diagram below (NOT to scale):

圖片參考：http://i1191.photobucket.com/albums/z467/robert1973/May11/Crazygeom3.jpg


Let BM = x, then AM = 9 - x. Similarly let NC = y, then AN = 7 - y

For the angles, we let ∠BAD = ∠CAD = θ, then:∠DMN = ∠DNM = θ (∠s in the same segment)∠BDM = ∠CDN = θ (∠s in alt. segment)Finally let ∠ABC = α and ∠ACB = βUsing cosine law, we have: 82 = 72 + 92 - 2 x 7 x 9 cos ∠A126 cos ∠A = 66cos 2θ = 11/212 cos2 θ - 1 = 11/21cos2 θ = 16/21 and hence sin2 θ = 5/21cos θ = √(16/21) and sin θ = √(5/21)Similarly,cos α = 2/3 and then sin α = (√5)/3cos β = 2/7 and then sin β = (3√5)/7Now, with ∠DMN = ∠DNM = ∠BDM = ∠CDN = θ, we have MN//BC (converse of alt. ∠s) Using intercept theorem, x/AB = y/AC = 1 - MN/BCx/9 = y/7 = 1 - MN/8So x = 9(1 - MN/8) ... (1) and y = 7(1 - MN/8) ... (2)Consider △CDN, ∠CND = 180 - (θ + β). Applying sine law: y/sin θ = CD/sin [180 - (θ + β)]CD = y sin (θ + β)/sin θConsider △BDM, ∠BMD = 180 - (θ + α). Applying sine law:x/sin θ = BD/sin [180 - (θ + α)]BD = x sin (θ + α)/sin θWith BD + BC = 8:y sin (θ + β)/sin θ + x sin (θ + α)/sin θ = 8(1/sin θ) [y sin (θ + β) + x sin (θ + α)] = 8√(21/5) [9(1 - MN/8) sin θ cos β + 9(1 - MN/8) cos θ sin β + 7(1 - MN/8) sin θ cos α + 7(1 - MN/8) cos θ sin α] = 8 (from (1) and (2))(1 - MN/8) √(21/5) (9 sin θ cos β + 9 cos θ sin β + 7 sin θ cos α + 7 cos θ sin α) = 8(1 - MN/8) √(21/5) [9 √(5/21) (2/7) + 9 √(16/21) (3√5)/7 + 7 √(5/21) (2/3) + 7 √(16/21) (√5)/3] = 8(1 - MN/8) (18/7 + 108/7 + 14/3 + 28/3) = 832(1 - MN/8) = 81 - MN/8 = 1/4MN/8 = 3/4MN = 6

圖片參考：http://imgcld.yimg.com/8/n/HA00400509/o/701105280086813873452950.jpg


圖片參考：http://imgcld.yimg.com/8/n/HA00400509/o/701105280086813873452961.jpg

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2011-05-28 22:33:43 補充：
http://imageshack.us/photo/my-images/28/113it.gif/