Given ray AE∥ ray BD. ray AF is bisector of ∠ EAB and ray BC is the bisector of ∠ Abd.
To find: line AF ∥ line BC
∠ EAB = 2x (ray AF bisector ∠ EAB)
When a line, shape, or angle inti two exactly equal parts is called bisector.
∠ ABD = 2y (ray BC bisector ∠ ABD)
Ray AE ∥ ray BD and Ab is transversal.
∠ EAD ≅ ∠ ABD (alternate angle) two angle formed when a line crosses two other lines, that lie on opposite side of the transversal line and on opposite relative sides of the other lines. If the two lines crossed are parallel, the alternate angles are equal.)
2x = 2y
X = y
∠ FAB∠ ABC
But they form a pair of alternate angle that are congruent.
∴ line AF ∥ line BC (hence proved)
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