Prove that in a regular hexagon ABCDEF, the diagonal AC divides it into 2 figures, the areas of which

Prove that in a regular hexagon ABCDEF, the diagonal AC divides it into 2 figures, the areas of which are proportional to the numbers 1: 5

Let the side length of a regular hexagon be X cm.

Let’s define the area of a regular hexagon, which consists of six areas of regular triangles AOB.

S = 6 * Saov = 6 * X * X * Sin60 / 2 = 6 * X2 * √3 / 4 = 3 * X2 * √2 / 2 cm2.

Determine the inner corner of a regular hexagon. Angle ABC = (6 – 2) * 180/6 = 720/6 = 120.

Determine the area of the triangle ABC.

Savs = AB * BC * Sin120 / 2 = X ^ 2 * √3 / 4.

Sdeck = S – Sас = (3 * X ^ 2 * √3 / 2) – X ^ 2 * √3 / 4 = 5 * √3 / 4 cm2.

Then Sас / Sсdec = (X ^ 2 * √3 / 4) / (5 * X ^ 2 * √3 / 4) = 1/5, which is what was required to prove.