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Proposed ProblemSee also: Complete Problem 281, Collection of Geometry ProblemsLevel: High School, SAT Prep, College geometry
Wonder if this problem has purely geometry proof. I solved it using trigonometry.
Extend AD, from C draw a segment,intersect with Extended AD at H, CH= AC, tri ACH is isosceles;connect BH;<BDC = 150-alfa, so <HDC = 90-alfa, so <BDH =60;because <ACD + <DCH + 4 alfa =180 =90-3alfa +<DCH + 4alfa ; so <DCH =90-alfa;so DHC is isosceles;So AC=CH=DH; so DH =DB, so BDH is isosceles; and <BDH is 60, so Tri BDH is equilateral tri,;so BH=DH=HC ; so use H as center, radius =HC, make a circle connect BDC. because <BDH=60, so X=30
Extend AD to E such that AC = CE< EDC = 90 - @ and < AEC = 2@ so DE = ECNow < BDE = 60 so BDE is equilateral and E is the centre of BCD. Hence x = 1/2 < BED = 1/2 X 60 = 30We need to observe that x = 30 irrespective of what value @ holdsSumith PeirisMoratuwaSri Lanka