Online Geometry theorems, problems, solutions, and related topics.
See complete Problem 9 at:www.gogeometry.com/problem/problem009.htmTriangle, Angles, Congruence. Level: High School, SAT Prep, College geometryPost your solutions or ideas in the comments.
I am brazilian, sorry for the bad english. Let E be the intersection of AC and BD. For triangle AED, angle(AEB) = x + 60. For triangle EBC, angle(ECB) = angle(AEB) – angle(EBC) = x + 60 – (x + 30) = 30. So angle(ADB) = 2.angle(ACB), points A, B and C belong to a same circle with center D, AD = DC, triangle ADC is isosceles and x = 10.
The solution is uploaded to the following link https://docs.google.com/open?id=0B6XXCq92fLJJdW5xcjU2OHVxbGM
As Nilton has pointed out: angle(ADB) = 2* angle(ACB) AND AD = BD which implies that D lies on the perpendicular bisector of AB which, in turn, means that D is the circumcentre of Tr.ABC; hence etc.
To Ajit: You're right, the condition AD = BD should be mentioned explicitly, though it is said in the enunciate. Thanks for your help.
Let perpendıcular bısector of AB (passing through D) intersect AC at E; <ADE=30, so <CED=30+x=<CBD and BEDC is cyclic, so <DBE=<DCE=10. As, by symmetry <DAE=<DBE, our answer is <DAE=10.Best regards,Stan Fulger
Triangle ABD is equilateral; let the perpendicular bisector of AB intersect AB at E and, clearly <CED=30+x=<CBD, so BEDC is cyclic and <DBE=<DCE=10 ( 1 ). Since by symmetry <DAE=<DBE, we have our answer, x=10.Best regards,Stan Fulger
Extend BD to E such that BD = DEThen < EAC = < EBC = x+30 and so ABCE is cyclic with D as centreHence Tr. ADC is isoceles and x=10Sumith PeirisMoratuwaSri Lanka
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