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As shown in the figure, ⊙ O's chord AB and AC have an included angle of 50 °, Mn is the midpoint of arc AB and arc AC respectively, and OM and on intersect AB and AC at points E and f respectively, then the degree of ∠ mon is () A. 110°B. 120°C. 130°D. 100°
∵ m and N are the middle points of arc AB and arc AC, respectively, ∵ of ⊥ AC, OE ⊥ AB, ∵ ofa = ∵ OEA = 90 °, ∵ in quadrilateral oeaf, ∵ mon = 360 ° - ∵ ofa - ∵ OEA - ∵ a = 360 ° - 90 ° - 90 ° - 50 ° = 130 °
As shown in the figure, OA, OB and OC are the three radii of the center point O of the circle, m and N are the two points on oaob, and am = 2om, BN = 2on, MC = NC. Prove: chord AC = chord BC
Because OA, ob is the radius of the center point O of the circle
So OA = ob,
Because am = 2om, BN = 2on
So om = 1 / 3oa, on = 1 / 3ob
So om = on
Because MC = NC, OC = OC
So △ OMC is all equal to △ onc
Therefore, AOC = BOC, because OA = ob, OC = OC,
So △ AOC is equal to △ BOC
So, string AC = string BC
Given a point a in circle O, (1) draw string Mn through point a, so that point a is the midpoint of string Mn; (2) draw three strings arbitrarily through point a, and try to compare these three strings
A point a in circle O is known,
(1) Draw string Mn through point a, so that point a is the midpoint of string Mn;
(2) After point a, draw three strings at will. Try to compare the length of these three strings with that of string Mn. What conclusion can you find?
This is a geometric problem. It is obvious that these three strings are longer than the string Mn, and the larger the angle with the string Mn, the longer the string. The longest is the string with an angle of 90 ° with the string Mn, which is the diameter of the circle!
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