University:
High School
Course:
High School Geometry
Academic year:

2020
Views:

391

Pages:

28
Author:

LTHrg

Proof of the Triangle Side Splitter Theorem Opening Exercise: We are going to construct a proof designed to demonstrate the following theorem: Side Splitter Theorem: A line segment splits two sides of a triangle proportionally if and only if it is parallel to the third side. If DE || AB, then ΔABC ~ ΔDEC triangle ABC with a line segment DE parallel to AB, intersecting sides AC and BC at points D and E respectively. Points W and Z are marked on line DE and AB respectively. Prove the theorem: Given: DE || AB Prove: ΔABC ~ ΔDEC Statements Reasons 1. DE AB 2. ∠1 ≅ ∠2 2. If the lines are parallel, then the corresponding ∠s are ≅ 3. ∠3 ≅ ∠4 3. If the lines are parallel, then the corresponding ∠s are ≅ 4. ΔABC ~ ΔDEC 4. AA Similarity What similarity transformation maps the smaller triangle onto the larger triangle? We have now proved the Triangle Side Splitter Theorem, which states: If a segment is parallel to a side of a triangle and intersects the other two sides, then this segment divides those two sides proportionally. 1 If DE || AB, then ΔABC ~ ΔDEC triangle ABC with a line segment DE parallel to AB, intersecting sides AC and BC at points D and E respectively. Points W and Y are marked on line DE and AB respectively. The following proportions exist: upper left side / lower left side = upper right side / lower right side → W / X = Y / Z upper left side / whole left side = upper right side / whole right side → W / W+X = Y / Y+Z lower left side / whole left side = lower right side / whole right side → X / W+X = Z / Y+Z Properties of Similar Triangles Opening Exercise: Given ΔABC ~ ΔA'B'C' pictured to the right: two similar triangles, ΔABC and ΔA'B'C'. ΔABC has sides 9, 12, and 15. ΔA'B'C' has sides 3, 4, and 5. Find the lengths of the missing sides. In ΔABC: a² + b² = c² x² + 9² = 15² x² + 81 = 225 x² = 144 x = 12 In ΔA'B'C': 9/15 = y/5 45/15 = 15y/15 3 = y Find the perimeter of the triangles. Large: P = 9 + 12 + 15 P = 36 Small: P = 3 + 4 + 5 P = 12 P = 3x more than the smaller Find the area of the triangles. Large: A = 1/2 * b * h A = 1/2 * 12 * 9 A = 54 Small: A = 1/2 * b * h A = 1/2 * 3 * 4 A = 6 A = 9x more than the smaller What is the ratio of the sides of the triangles? Sides 3:1 The sides are in proportion with a constant scale factor of 3. What is the ratio of the perimeters of the triangles? Perimeter 3:1 The perimeters are in the same proportion 1 as the sides, 3x larger. 1. What is the ratio of the areas of the triangles? Areas 9:1 The areas are 9x larger from the large Δ to the small Δ. Let's test these relationships with other similar figures. Given the similar polygons pictured below, find: Ratio of the sides: 24/36 = 2/3 10/15 = 2/3 26/39 = 2/3 Ratio of the perimeters: Big: P = 15 + 36 + 39 = 90 Small: P = 10 + 14 + 16 = 40 90/40 = 9/4 Ratio of the areas: Big: A = 1/2 * b * h = 1/2 * 15 * 36 = 270 Small: A = 1/2 * 10 * 24 = 120 270/120 = 9/4 Given the similar rectangles pictured to the right, find: Ratio of the sides: 20/10 = 2 50/25 = 2 Ratio of the perimeters: Big: P = 2(20) + 2(50) = 40 + 100 = 140 Small: P = 2(10) + 2(25) = 20 + 50 = 70 140/70 = 2 Ratio of the areas: Big: A = b * h = 20 * 50 = 1000 Small: A = 10 * 25 = 250 1000/250 = 4