In figure 2.21, if ray BA || ray DE, ∠ c = 50° and ∠ D = 100°. Find the measure of ∠ ABC.

(Hint: Draw a line passing through point C and parallel to line AB.).

Given: ray ∠ BA ∥ ∠ DE and ∠ c = 50°, ∠ D = 100°.

To find: ∠ ABC.

Construction: Extend AB such that A-B-F-G.

BA ∥ DE (given)

AG∥ DE(construction) and DC is transversal.

∠ D ≅ ∠ GFC ((corresponding angle theorem) if two parallel line are cut by a transversal, then the pairs of corresponding angle are congruent).

∠d = 100° (given)

So that ∠ GFC = 100°

∠ GFC + ∠ BFC = 180(linear pair angle

∠ GFC = 100 (proved above)

∠ GFC + ∠ BFC = 180°

∠ 100 + ∠ BFC = 180°

∴ ∠ BFC = 100-180°

∠ BFC = 80°

In Δ BFC

∠ BFC + ∠ c + ∠ FBC = 180° (sum of angle of Δ)

∠ 80° + 50° + ∠ FBC = 180° (already given value above ∠ BFC and ∠ c)

∠ FBC = 180° -50°

∠ FBC = 130°

∠ ABC + ∠ FBC = 180°

∠ ABC + 50 ° = 180°

∠ ABC = 180-50°

∠ ABC = 130.