Given Problem:
In the given geometry problem, here’s the information and figure provided:
- (\overline{EK}) is parallel to (\overline{DC}).
- (\overline{AE}) is equal to (\overline{EB}).
- (DE = 8) cm, (DB = 6) cm.
- (\overline{BC} = 12) cm.
- (\overline{EF} = x).
We are asked to find the value of (x).
Solution:
Let’s analyze the situation step-by-step:
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Since (\overline{EK} \parallel \overline{DC}), triangles (\triangle AEK) and (\triangle ADC) are similar by the Basic Proportionality Theorem (Thales’ theorem).
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Given (\overline{AE} = \overline{EB}), it implies point (E) divides (\overline{AB}) in the ratio 1:1, hence (\overline{AE} = \overline{EB}).
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Using the properties of similar triangles, the ratio of the sides of (\triangle AEK) and (\triangle ADC) will be equal.
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We need to find out the ratio in which (F) divides (\overline{BD}).
Let’s consider the triangle (\triangle BDC):
- (DB = 6) cm.
- (DE = 8) cm.
- (\overline{BC} = 12) cm.
The point (F) divides (BD) in such a way that:
[ BF = \frac{DE}{DE + DB} \times BC = \frac{8}{8+6} \times 12 = \frac{8}{14} \times 12 = \frac{8}{14} \times 12 = \frac{96}{14} = \frac{48}{7} \approx 6.857 ]
Now, assuming the ratio (\frac{DF}{BF} = 1), as (\overline{EF}) is parallel to (\overline{DC}).
Given:
[ DK = EF ]
So, (EK = x \times \frac{1}{2} \approx x = 2 )
Therefore, (\overline{EF} = x = 2)
Final Answer:
The value of (x) is ( \boxed{2} ).
So the correct answer is:
[ \boxed{2} ]