Question

In the given figure, if ray BA || ray DE, $\angle$C = 50^° and $\angle$D = 100^° . Find the measure of $\angle$ABC.

Answer 1

Draw a line XY passing through point C and parallel to AB.
We have, AB || DE and AB || 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.
∴ DE || XY.
Since DE || XY and DC is a transversal intersecting them at D and C, then
∠EDC + ∠YCD = 180^∘ (Pairs of interior angles on the same side of transversal are supplementary)
⇒100^∘ + ∠YCD = 180^∘
∠YCD = 180^∘ − 100^∘ = 80^∘
Since, sum of all the angles on a straight line at a point is 180^∘, then
∠XCB + ∠BCD + ∠YCD = 180^∘
⇒∠XCB + 50^∘ + 80^∘ = 180^∘
⇒∠XCB + 130^∘ = 180^∘
⇒∠XCB = 180^∘ − 130^∘ = 50^∘
Since, AB || XY and BC is a transversal intersecting them at B and C, then
∠ABC + ∠XCB = 180^∘ (Pairs of interior angles on the same side of transversal are supplementary)
⇒∠ABC + 50^∘ = 180^∘
⇒∠ABC = 180^∘ − 50^∘ = 130^∘