Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 760 details.
http://img705.imageshack.us/img705/9923/problem760.pngConnect lines per sketch.Note that F and H are midpoints of AC and DCIn right triangle BFC we have CF^2=CB. CH= 1…. ( relation in right triangle)In right triangle BEC we have CE^2=x^2=CD.CB=2CH.CB=2So x=√ 2
x^2=CE^2=CD*CB & CD/CA=(CA/2)/CB or CA^2=2CD*CB or x^2=CA^2/2=(4/2)=2 hence x=√2
Let AC intersect circle at F.∠BFC is 90° (angle in semicircle).So BF is ⊥ bisector of AC, and so CF = (1/2) AC = 1Also A, B, D, E are concyclic (∵∠AFB = ∠ADB = 90°)Now In the right ∆BEC, ED ⊥ BC. So x² = CE² = CD.CB = CF. CA = (1)(2) = 2, x = √2
I have two different solutions for this problem, but i dont have time to post it (video is comming soon). Anyways, my post have another intention, i want to comment that if triangle is obtuse, the problem is absurd. In my opinion is necessary tell that ABC is an acute triangle.
Extend CB to F such that CB = BF. Then < CAF = 90 and so 2^2 = CD.CFSimilarly since < CAB = 90 x^2 = CD.CBComparing these 2 equations and since CF = 2. CB we have 2^2 = 4 = 2.x^2So x = sqrt 2Sumith PeirisMoratuwaSri Lanka