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# 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.

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## Answer to a math question 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.

Santino
4.5
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.

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