In the diagram EFGH is a cyclic quadrilateral i... - JAMB Mathematics 1999 Question

< EFH = < EGH = 34^o (angles n the same segment)

< GHF = < GEF = 42^o(angles in the same segment)

< FOG = 42 + 34 = 76(exterior angle)

< FOG = < EOH = 76(vertically opposite angle)

< EDO = 90^o, < DOE = \(\frac{76}{2}\) = 38^o

< HEG = 90^o - 38^o = 52^o