i am lost on this one
TAKE Three points K,L,M on sides BC,CA,AB of the triangle ABC. Prove that
at least one of the triangles AML,BKM,CLK has area less than or equal to 1/4 area of triangle ABC
To prove that at least one of the triangles AML, BKM, and CLK has an area less than or equal to one-fourth of the area of triangle ABC, you can use the concept of the area of triangles and the triangle inequality.
Let's start by considering the area of a triangle. The area of a triangle can be calculated using the formula:
Area = (base * height) / 2
Now let's consider triangle ABC. Its area is denoted as Area(ABC), and you want to prove that at least one of the triangles AML, BKM, and CLK has an area less than or equal to one-fourth of Area(ABC), which can be expressed as Area(ABC)/4.
To prove this, you can divide it into cases, considering each of the three sub-triangles separately.
1. Case 1: Triangle AML
Let's assume that the area of triangle AML, denoted as Area(AML), is greater than one-fourth of Area(ABC) (i.e., Area(AML) > Area(ABC)/4).
Now, consider the heights of triangle AML and triangle ABC. The height of triangle AML is h(AML), and the height of triangle ABC is h(ABC).
Since triangle AML is a sub-triangle of triangle ABC, its height h(AML) must be less than or equal to the height of triangle ABC, i.e., h(AML) ≤ h(ABC).
Using this information, we can compare the areas of triangle AML and triangle ABC using the formula mentioned earlier:
Area(AML) = (base(AML) * h(AML)) / 2
Area(ABC) = (base(ABC) * h(ABC)) / 2
Since h(AML) ≤ h(ABC), and Area(AML) > Area(ABC)/4, we can conclude that:
(base(AML) * h(AML)) / 2 > (base(ABC) * h(ABC)) / 8
Next, consider that base(AML) ≤ base(ABC) because AML is a sub-triangle of ABC. Therefore, (base(AML) * h(AML)) / 2 ≤ (base(ABC) * h(AML)) / 2.
Combining the above inequalities, we have:
(base(ABC) * h(AML)) / 2 > (base(ABC) * h(ABC)) / 8
Canceling the common terms, we get:
h(AML) > h(ABC) / 4
This implies that the height of triangle AML is greater than one-fourth of the height of triangle ABC.
However, this contradicts our initial assumption that Area(AML) > Area(ABC)/4. Hence, our assumption was incorrect. Therefore, the area of triangle AML must be less than or equal to one-fourth of the area of triangle ABC.
By similar reasoning, you can prove that either BKM or CLK must also have an area less than or equal to one-fourth of the area of triangle ABC.
Hence, it has been proven that at least one of the triangles AML, BKM, or CLK has an area less than or equal to one-fourth of the area of triangle ABC.