Complete the following theorem: If a line divides two sides of a triangle in the same proportion, then the line is .. - NSC Technical Mathematics - Question 9 - 2019 - Paper 2

Using the triangles ODF and OBC, since DF || BC, we have:

$\frac{BC}{DF} = \frac{OB}{OD} = \frac{5}{3}$ Thus, the ratio of BC : DF is 5 : 3.

Given that OD = OE and OB = OG, we can find EG:

$EG = OE - EG$ Substituting the values:

$EG = 6 - 3.6 = 2.4$ Therefore, the length of EG is 2.4 units.

To find the areas, we use:

Area of triangle = \frac{1}{2} \times base \times height.

( Area (\Delta OBG) = \frac{1}{2} (6)(3.6) = 10.8 ) ( Area (\Delta ODE) = \frac{1}{2} (8)(2) = 8 )

Thus,

$\frac{Area(\Delta OBG)}{Area(\Delta ODE)} = \frac{10.8}{8} = 1.35$ The numerical value of Area Δ OBG / Area Δ ODE is approximately 1.35.

In triangles DOE and BOG, since DF || BC, angles DOF and BGO are equal by the Alternate Interior Angles Theorem. Additionally, angles ODE and OBG are also equal because they are both right angles. Therefore, by AA similarity, we have that Δ DOE || Δ BOG.