In the diagram, EFGH is a cyclic quadrilateral in which EH || FG, EG, and FH are chords. If < FHG = 42º and < EFH = 34º, calculate < HEG.
34o
42o
52o
76o
Correct answer is C
< EFH = < EGH = 34º (angles n the same segment)
< GHF = < GEF = 42º(angles in the same segment)
< FOG = 42 + 34 = 76º (exterior angle)
< FOG = < EOH = 76º (vertically opposite angle)
< EDO = 90º, < DOE = \(\frac{76}{2}\) = 38º
< HEG = 90º - 38º = 52º
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