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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to open the complete problem 867.
Draw CF perpendicular to Ad extended. /_FCD=/_DBE=/_DAC; hence FC is tangent to the circum-circle of tr. ADC and thus CF²=BE²=1*(x+2) since FD=DE=1. Now AB²=BE²+x², AB=2*BD and BD²=1+BE² give us: x²+2+x=4(x+2)+4 which in turn gives x=5
Let bisector of < ABC meet AD at F and AC at G. Let < CAD = < EBD = @ABEG is cyclic with diameter AB so < FBE = @. Hence FE = 1 and AF = x-1Since BF is an angle bisector of Tr. ABD, AB/BD = (x-1)/2 from which x = 5 since AB/BD = 2Sumith PeirisMoratuwaSri Lanka