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Official PYQ · 2018QGA-9MCQ2 marks

GATE CS 2018 QGA-9 — General Aptitude

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In the figure below, ∠DEC+∠BFC\angle DEC + \angle BFC is equal to _____

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Answer: A

Let the interior angles of the quadrilateral ABCDABCD be denoted as A,B,C,A, B, C, and DD. The sum of the interior angles of a quadrilateral is 360∘360^\circ:

A+B+C+D=360∘  ⟹  B+D=360∘−(A+C)A + B + C + D = 360^\circ \implies B + D = 360^\circ - (A + C)

From the geometry of the figure, we can observe the following triangles:

  1. In △ADE\triangle ADE, the sum of angles gives: ∠DEC=180∘−(A+D)\angle DEC = 180^\circ - (A + D)
  2. In △ABF\triangle ABF, the sum of angles gives: ∠BFC=180∘−(A+B)\angle BFC = 180^\circ - (A + B)

Adding these two equations:

∠DEC+∠BFC=360∘−2A−(B+D)\angle DEC + \angle BFC = 360^\circ - 2A - (B + D)

Substituting B+D=360∘−(A+C)B + D = 360^\circ - (A + C) into the equation:

∠DEC+∠BFC=360∘−2A−(360∘−A−C)=C−A\angle DEC + \angle BFC = 360^\circ - 2A - (360^\circ - A - C) = C - A

Thus, ∠DEC+∠BFC=∠BCD−∠BAD\angle DEC + \angle BFC = \angle BCD - \angle BAD.

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