NEED ASAP For a sewing project, Tanya cut isosceles triangles from a striped piece of material where the stripes are parallel. The vertex angle of the isosceles triangle is 50° and overline BC is parallel to the base.

What is the measure of BCE as shown in the diagram? Show your work and
explain your reasoning.

∠BCE is 115°

Step-by-step explanation:

The given parameters are;

The shape Tanya cut from the striped piece = Isosceles triangles

The measure of the vertex angle of the isosceles triangle = 50°

The segment is parallel to the base

Therefore, we have;

∠GAF + ∠AGF + ∠AFG = 180°; Sum of the interior angles of a triangle

∴ ∠AGF + ∠AFG =  180° - ∠GAF =  180° - 50° = 130°

∠AGF = ∠AFG;  Base angles of an isosceles triangle

∴ ∠AGF + ∠AFG = ∠AGF + ∠AGF = 2 × ∠AGF = 130°

∠AGF = 130°/2 = 65°

∠AGF = 65°

∠AGF ≅ ∠ACB, corresponding angles of two parallel lines, and cut by a transversal

∠AGF = ∠ACB = 65°, definition of congruency

∠ACB + ∠BCE = 180° Sum of angles on a (straight) line

∠BCE = 180° - ∠ACB =  180° - 65° = 115°

∠BCE = 115°

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