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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 718 details.
Extend CDO' to cut the circle O' at E. Join AE, BE. Let angle AEB = yBy the result of Problem 717,y = 65 degBut AEBD is a cyclic quadrilateral.So x + y = 180 degHence x = 115 deg
By applying the formula that angle bda=90+bca/2 therefore half of 50=25 and by adding 90 we get 90+25=115therefore measure of angle bda =115
x = angle subtended by major arc AB of circle(O')at D= half the ∠ subtended by arc AB of circle (O')at center O'.= (1/2)major∠AO'B= (1/2)(360°-minor∠AO'B)= (1/2)[360°-(180°-∠ACB)] (∵ AO'BC is a cyclic quadrilateral)= (1/2)(180°+∠ACB)= (1/2)(180°+50°)= (1/2)(230°)= 115°