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Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Since ∠OED=∠OHA, so OEDH concyclic. Thus ∠EOH=90°. Also, OE=OH=5. Hence, EH=5√2.
Rotate figure by 90° counter-clockwise around O. A goes to D and H goes to E. Hence OH=OE, EOH=90°, EH=5√2
Tr.s DOE and AOH are congruent SAS and so OH = OESo < AOH = < DOE = 67.5 and since < AOD = 90, HOE = 90Hence Tr. HOE is right isoceles and x = 5sqrt2Sumith PeirisMoratuwaSri Lanka