[email protected] wrote:Hi Experts,
Please help me with the tricks being used in this problem.
Thanks
Since AB is parallel to EH, AG serves as the transversal, and therefore if angle EGC = 70°, angle BAC is also equal to 70°. In triangle ABC, AB = BC, hence BCA also equals 70°.
Also, BD is parallel to FH and with DF serving as transversal, the co-interior angles at D and F are supplementary, in other words, angle at D will be equal to the sum of equal angles FEH and FHE in the isosceles triangle EFH.
Focus on the interior angles in quadrilateral CDEG now, C = G = 70°. Try plugging in the answers from here onward.
(C) If D = 75°, angle at F is 105°, and this will result fractional angle at E. Not fitting, eliminate C and so eliminate all odd numbered choices, A and E also eliminated, as those would result in a fractional angle at E.
Try (D) 80° as (B) 70° sounds like a trap answer. If D = 80°, angle at F is 100°, angles FEH and FHE would equal 40° each, or angle DEG will equal 140°. This satisfies the angle sum property of a quadrilateral; hence [spoiler]
(D)[/spoiler] is the right answer.
