So what can you do? Recall that with TU ‖ QR , you can show that Δ QRS∼ Δ TUS. This means that you cannot apply Theorem 57 to this situation. Notice that TU , x, is not one of the segments on either side that TU intersects. Theorem 57 (Side‐Splitter Theorem): If a line is parallel to one side of a triangle and intersects the other two sides, it divides those sides proportionally.įigure 2 Using the Side‐Splitter Theorem.īecause DE ‖ AC in Δ ABC by Theorem 57, you get With this replacement, you get the following proportion. Now use Property 4, the Denominator Subtracion Property.īut AB–DB = AD, and BC–BE = CE ( Segment Addition Postulate). You can eventually prove that Δ ABC∼ Δ DBE using the AA Similarity Postulate. Because the ratios of corresponding sides of similar polygons are equal, you can show that Summary of Coordinate Geometry FormulasĬonsider Figure 1 of Δ ABC with line l parallel to AC and intersecting the other two sides at D and E.įigure 1 Deriving the Side‐Splitter Theorem.Slopes: Parallel and Perpendicular Lines.Similar Triangles: Perimeters and Areas.
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