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Question
In the given figure, if ray BA $$\parallel$$ ray DE, $$\angle C = 50^\circ$$ and $$\angle D = 100^\circ.$$ Find the measure of $$\angle ABC.$$
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Solution
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Draw a line XY passing through point C and parallel to AB. We have, $$AB \parallel DE$$ and $$AB \parallel XY.$$ It
is known that, if two lines in a plane are parallel to a third line in
the plane, then those two lines are parallel to each other. $$\therefore DE \parallel XY.$$ Since $$DE \parallel XY$$ and DC is a transversal intersecting them at D and C, then $$\angle EDC + \angle YCD = 180^\circ$$ (Pair of interior angles on the same side of transversal are supplementary) $$\Rightarrow 100^\circ + \angle YCD = 180^\circ$$ $$\angle YCD = 180^\circ - 100^\circ = 80^\circ$$ Since, sum of all the angles on a straight line at a point is $$180^\circ,$$ then $$\angle XCB + \angle BCD + \angle YCD = 180^\circ$$ $$\Rightarrow \angle XCB + 50^\circ + 80^\circ = 180^\circ$$ $$\Rightarrow \angle XCB + 130^\circ = 180^\circ$$ $$\Rightarrow \angle XCB = 180^\circ - 130^\circ = 50^\circ$$ Since, $$AB \parallel XY$$ and BC is a transversal intersecting them at B and C, then $$\angle ABC + \angle XCB = 180^\circ$$ (Pairs of interior angles on the same side of transversal are supplementary) $$\Rightarrow \angle ABC + 50^\circ = 180^\circ$$ $$\Rightarrow \angle ABC = 180^\circ - 50^\circ = 130^\circ$$
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In the given figure, if ray BA || ray DE, C = 50Ā° and D = 100Ā° . Find the measure of ABC.