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Geometry ProblemProblem submitted by Charles T.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 832.
http://img51.imageshack.us/img51/5048/problem832.pngFrom D draw DE//BC and DE=BC=ADBCDE is a parallelogram => BE //DC => BE is angle bisector of angle ABDSo ED=AD=AE => ADE is a quadrilateral triangle and ∠ ADE=60So ∠BDC=∠BDE=70-60=10
AnonymousWithin angle ABD build angle equilateral APD , BP perpendicular bisector for the segment AD . BP parallel with DC , DP=BC than BPDC trapezoid or parallelogram, BPD=150grade when ABC <90 we have parallelogram, and angle DPC is 10 grade .When 90<ABC<180 it becomes that DPC=130 gradeErina New Jersey
Complete the rectangle ADCE. Join BE. Triangle BEC is equilateral.Since triangles BEA,BCD are congruent,angles ABE,DBC are equal each equal to (60 - 40)/2 = 10 degrees.
Construim paralelogramul ADMB si vom triunghiul echilateral BMC deoarece triunghiurile ABC=BDM si BCD=MCD=>m(<CBD)=m(<MBD)-m(<CBD)=70-60=10
On the angle bisector of B mark E such that BE = CD. But BE is also parallel to CD hence BECD is a parelleligram and DE = and is oarellel to BC = AE SASHence Tr. AED is equilateral and x = < EDB = 10Sumith PeirisMoratuwaSri Lanka