Proposed Problem
Click the figure below to see the complete problem 405 about Quadrilateral, 60, 75, and 135 degrees, Midpoint.
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Complete Problem 405
Level: High School, SAT Prep, College geometry
Sunday, December 20, 2009
Problem 405. Quadrilateral, 60, 75, and 135 degrees, Midpoint
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http://geometri-problemleri.blogspot.com/2009/12/problem-55-ve-cozumu.html
ReplyDeleteA slightly different way with one application of the sine rule gives us the following solution: Join A to C & E. Tr. ACD is equilateral since /_D=60 and CD=AD=2. Thus, /_CAD=60 & /_BAC=15,/_AEC=90, AE=√3 while /_ACE=60. Further, quad ABCE is concyclic & thus /_ABE = /_ACE=60 and finally tr. ABE gives, x/√3 = sin(45)/sin(60) resulting in x=√2
ReplyDeleteAjit
join A to C, draw middle line EF = 1 ( F on AC ), join B to F. Win right isosceles tr BFE, BF = Ef = 1, angBFE=90
ReplyDelete(BF is median for tr ABC)Pythagore theorem => x = v2
Purely Plane Geometry (No Need to Use Trigonometry):
ReplyDeleteConnect AC and AE, Extend Ray AE and Ray BC till they intersect (call the intersection F).
Since AD=CD=2 and Angle ADC=60 Degrees, ACD is equilateral. (AC=2, Angle ACD=60 Degrees.) AE=√3.
In ACE and ADE, AC=AD, Angle ACD=Angle ACD, CE=ED (given midpoint), so ACE congruent to ADE, Angle CAE=30 Degrees. Angle BAC=Angle BAD-Angle CAD=(75-60)Degrees=15 Degrees. Angle BAE=Angle BAC+Angle CAE=(15+30)Degrees=45 Degrees. So it's revealed ABCE is cyclic quadrilateral (Angle BCE+Angle BAE)=(135+45)Degrees=180 Degrees.
According to Ptolemy's Theorem, AC*BE=BC*AE+AB*CE. So we have 2x=√3*BC+AB. Angle BAE=Angle CFE=Angle FCE=45 Degrees. AF=√3+1. BF=AB=(√6+√2)/2, BC=(√6-√2)/2. Calculate:
2x=√3*(√6-√2)/2+(√6+√2)/2=2√2,
Therefore, x=√2.