In the given figure, two straight lines E-A-C-F and R-A-B-P intersect at A above the line M-B-C-N (M, B, C, N collinear). If m∠BCF = 110°, m∠MBP = x°, and m∠BAE = 2x°, find the value of m∠ABC.
70°
Since B, C, F are such that F lies on line AC extended beyond C, ∠BCF and ∠ACB are a linear pair: ∠ACB = 180 - 110 = 70°.
Since M, B, C, N are collinear and P, A, B lie on another straight line, ∠MBP is vertically opposite to ∠ABC, so ∠MBP = ∠ABC = x°.
Since E, A, C are collinear, ∠BAE and ∠BAC are a linear pair: ∠BAC = 180 - 2x.
By angle sum property of triangle ABC: ∠BAC + ∠ABC + ∠ACB = 180
(180 - 2x) + x + 70 = 180 ⇒ 250 - x = 180 ⇒ x = 70.
So m∠ABC = 70°. Answer: (C).
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