Show that by joining the midpoints of an equilateral triangle, we divide it into 4 identical equilateral triangles.
Given: Let ABC be an equilateral triangle with side length s. D, E, F are midpoints of sides BC, CA, and AB respectively.
Construction: Join D, E, F to create four smaller triangles: AEF, BDF, CDE, and DEF.
Proof for Corner Triangles: For triangle AEF: AE = AF = s/2 (midpoints) Angle A = 60° (equilateral triangle ABC) ∴ Triangle AEF is equilateral with side s/2. Similarly, triangles BDF and CDE are equilateral with side s/2.
Proof for Central Triangle (DEF): By the midpoint theorem: DE || AB and DE = AB/2 = s/2 EF || BC and EF = s/2 FD || AC and FD = s/2 ∴ Triangle DEF is equilateral with side s/2.
Answer: All four triangles have side length s/2. ∴ All four triangles are identical equilateral triangles, each with area one-fourth of the original triangle.
Explanation
The chapter provides a hint that the corner triangles are isosceles. By the midpoint theorem, the line segment joining midpoints of two sides is parallel to the third side and half its length. This ensures all four triangles formed are equilateral with equal side lengths.
Solution Steps
Step 1: Let ABC be an equilateral triangle with midpoints D, E, F on sides BC, CA, AB.
Step 2: By the midpoint theorem, DE || AB and DE = AB/2 = s/2.
Step 3: Similarly, EF || BC and EF = s/2, and FD || AC and FD = s/2.
Step 4: Triangle DEF has all sides equal to s/2, so it is equilateral.
Step 5: Each corner triangle (like AEF) has two sides of length s/2 and included angle 60°, making it equilateral.
Step 6: All four triangles have side s/2, so they are identical equilateral triangles.