Proving Parallelograms: ABCDEFGH with IJK

[orangeSLICE]

Hello :D

ABCDEFGH is a cube with a side of 3cm. The points I and J are defined by:

vector BI = 1/5 of vector BF
vector CJ = 4/5 of vector CG

The plan (EIJ) cuts line (HG) into K.

1. Prove that line (EK) and line (FG) are parallele.

Any suggestions on how to go about this?

 

[VietDao29]

I really don't think the problem you posted is correct. Please re-check the problem.
Viet Dao,

 

[tacman]

To prove that lines (EK) and (FG) are parallel, we can use the properties of parallelograms. First, we know that the opposite sides of a parallelogram are equal in length. So, we can show that line (EK) and line (FG) have equal lengths.

To do this, we can use the given information that vector BI is 1/5 of vector BF and vector CJ is 4/5 of vector CG. Since BI and CJ are both diagonals of the cube, we can use the Pythagorean theorem to find their lengths.

Let's start with vector BI. Since it is 1/5 of vector BF, we can express it as BI = 1/5 * BF. Since BF is a side of the cube with a length of 3cm, we can substitute it into the equation and get BI = 1/5 * 3cm = 0.6cm.

Next, let's find the length of vector CJ. Since it is 4/5 of vector CG, we can express it as CJ = 4/5 * CG. Using the same logic as before, we can substitute the length of CG, which is also 3cm, and get CJ = 4/5 * 3cm = 2.4cm.

Now, we can use the Pythagorean theorem to find the lengths of lines (EK) and (FG). Let's start with line (EK). We know that the length of vector BI is 0.6cm and the length of vector CJ is 2.4cm. So, using the Pythagorean theorem, we can find the length of line (EK) as follows:

EK^2 = BI^2 + CJ^2
EK^2 = (0.6cm)^2 + (2.4cm)^2
EK^2 = 0.36cm^2 + 5.76cm^2
EK^2 = 6.12cm^2
EK = √6.12cm ≈ 2.47cm

Similarly, we can find the length of line (FG) using the same process:

FG^2 = BI^2 + CJ^2
FG^2 = (0.6cm)^2 + (2.4cm)^2
FG^2 = 0.36cm^2 + 5.76cm^