In rectangle ABCD, EF=2cm, DE=3cm, E is the intersection of diagonals, EF is perpendicular to AB. Find BF.

Here are the solutions for questions 6, 7, and 8 based on the provided image.

First, let's analyze the given information and the diagram:
The figure $ABCD$ is a rectangle.
$EF = 2$ cm.
$DE = 3$ cm.
$E$ is the intersection point of the diagonals $AC$ and $BD$.
$EF \perp AB$.

Properties of a rectangle:

• Diagonals are equal in length and bisect each other. Therefore, $AE = BE = CE = DE$. Since $DE = 3$ cm, we have $AE = BE = CE = 3$ cm.
• All angles are $90^\circ$. So, $\angle DAB = 90^\circ$.

Consider $\triangle ABD$:
$E$ is the midpoint of the hypotenuse $BD$ (since $BE = DE = 3$ cm).
$EF$ is drawn from $E$ such that $EF \perp AB$.
Since $AD \perp AB$ (because $ABCD$ is a rectangle), $EF$ must be parallel to $AD$.
By the midpoint theorem (or intercept theorem), if a line segment passes through the midpoint of one side of a triangle and is parallel to another side, it bisects the third side.
Thus, $F$ is the midpoint of $AB$.
Also, $EF = \frac{1}{2} AD$.
Given $EF = 2$ cm, we can find $AD$:
$2 = \frac{1}{2} AD$
$AD = 2 \times 2 = 4$ cm.
Since $ABCD$ is a rectangle, $BC = AD = 4$ cm.

Now we have:
$AD = 4$ cm
$BE = 3$ cm
$EF = 2$ cm
$F$ is the midpoint of $AB$.

6. $BF$ এর দৈর্ঘ্য কত সে.মি.? (What is the length of $BF$ in cm?)