Let G be the center of gravity of triangle ABC. We draw through A a parallel to BC on which we take a point D so that DG⊥BG. If the area of the quadrilateral AGBD is equal to s, show that AC·BD≥2·s.

To prove that AC · BD ≥ 2s, we need to use the fact that the area of triangle ABC is twice the area of triangle ADB.

Step 1: Draw a diagram to visualize the given information.

Step 2: Since DG ⊥ BG, triangle BDG is a right triangle. Let's call the length of DG as h.

Step 3: The area of triangle BDG can be calculated using the formula: Area(BDG) = (1/2) * BG * h.

Step 4: The area of triangle ADB can be calculated using the formula: Area(ADB) = (1/2) * AD * h.

Step 5: Since the area of AGBD is equal to s, we can express the area of triangle ADB as: Area(ADB) = s - (1/2) * BG * h.

Step 6: The area of triangle ABC is equal to twice the area of triangle ADB, so Area(ABC) = 2 * Area(ADB).

Step 7: Substitute the expressions for Area(ADB) and Area(ABC) from steps 5 and 6, respectively.

2 * (s - (1/2) * BG * h) = AC * BD.

Step 8: Simplify the expression and rearrange to get the desired inequality.

2s - BG * h = AC * BD.

Multiply both sides by 2:

4s - 2BG * h ≥ 2AC * BD.

Rearrange:

2AC * BD ≥ 4s - 2BG * h.

Since 4s - 2BG * h ≥ 0 (as areas cannot be negative), we can conclude that:

2AC * BD ≥ 2s.

Answer: AC · BD ≥ 2s.