Equilateral triangles are drawn on all the sides of a right triangle ABC. Prove that ar( ACF) = ar( BCE) + ar( ABD).
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See diagram. AC is the diagonal. Draw altitudes from vertices F, E, D onto the opposite sides.
Ar (Δ ACF) = 1/2 * FI * AC = 1/2 * √3/2 * AC * AC equilateral triangle.
similarly,
Ar (Δ BCE) = 1/2 * CH *BC = 1/2 * √3/2 BC * BC
Ar (Δ ABD) = 1/2 * DG * AB = 1/2 * √3/2 AB * AB
so LHS = √3/4 AC² = RHS = √3/4 * (BC²+AB²) as ABC is a right angle triangle.
Ar (Δ ACF) = 1/2 * FI * AC = 1/2 * √3/2 * AC * AC equilateral triangle.
similarly,
Ar (Δ BCE) = 1/2 * CH *BC = 1/2 * √3/2 BC * BC
Ar (Δ ABD) = 1/2 * DG * AB = 1/2 * √3/2 AB * AB
so LHS = √3/4 AC² = RHS = √3/4 * (BC²+AB²) as ABC is a right angle triangle.
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