We know that the sum of all angles of a triangle is 180°
Therefore, for the given △ABC, we can say that:
∠ABC + ∠BAC + ∠ACB = 180°
100° + 35° + ∠ACB = 180°
∠ACB = 180° –135°
∠ACB = 45°
∠C = 45°
On applying same steps for the △BCD, we get
∠BCD + ∠BDC + ∠CBD = 180°
45° + 90° + ∠CBD = 180° (∠ACB = ∠BCD and BD parallel to AC)
∠CBD = 180° – 135°
∠CBD = 45°
We know that the sides opposite to equal angles have equal length.
Thus, BD = DC
DC = 2 cm