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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 $$\angle ABC.$$

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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