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In the following diagram, everything occurs in a plane. AB and BC are line segments. DE is a line parallel to AB, and FG is a line parallel to BC. The distance between AB & DE is equal to the distance between BC & FG. Given the points A, B, and C, and the distance between AB & DE, what is the most efficient way to find the location of point I, the intersection of DE and FG?

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My initial approach was to calculate two points on the line DE, two points on the line FG, and find the intersection point of the two lines through those points. However, I was thinking, would the line HJ which passes though both B and I always cut the angle ABC exactly in half? If so, then perhaps a faster method of finding I is to calculate the angle ABI, the distance BI, and find I relative to B.

My geometry is a little rusty: although it seems to me like angle ABI should always be one half of angle ABC, I don't want to assume that without a little more rigor.

My question is: Does angle ABI always equal half of angle ABC, and/or, do you see a more efficient way of determining the location of point I given A, B, C, and the distance between AB & DE?

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Through B, draw PQ perpendicular to DE cutting it at H. Do the same thing for the line RKS.

(Right angle) + (BH = BK) + (BI is the common side) implies triangles BHI and BKI are congruent. [RHS]

Therefore, $\angle BIH = \angle BIK$ [corresponding measurements]

Thus, $\angle ABJ = \angle BIH = \angle BIK = \angle CBJ$ [corresponding angles of parallel lines]

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