Question

finding an unknown length
what is the length of $overline{bc}$?
from the markings on the diagram, we can tell e is the midpoint of $overline{bc}$ and $square$ is the midpoint of $overline{ac}$
we can apply the $square$ theorem: $ed = \frac{1}{2}ba$.
substituting in the expressions for the lengths and solving for $x$, we get $x = \square$
now, since $be = x$, then $bc = \square$
(options for $bc$: 2.5, 6, 10)
(diagram: triangle with points b, a (segment $4x - 6$), e (segment $be = x$), d (segment $ed = x + 2$), c; markings show $be = ec$ (single tick) and $ad = dc$ (double tick))

Explanation:

Step1: Identify Midpoints

From the diagram, E is midpoint of \( \overline{BC} \) (since \( BE = EC = x \)) and D is midpoint of \( \overline{AC} \) (since \( AD = DC \) from markings). So first blank: D.

Step2: Apply Midline Theorem

The theorem is Midline (or Midsegment) Theorem: \( ED=\frac{1}{2}BA \). So second blank: Midline (Midsegment).

Step3: Substitute and Solve for \( x \)

Given \( ED = x + 2 \), \( BA = 4x - 6 \). By midline theorem:
\( x + 2=\frac{1}{2}(4x - 6) \)
Multiply both sides by 2: \( 2(x + 2)=4x - 6 \)
Expand: \( 2x + 4 = 4x - 6 \)
Subtract \( 2x \): \( 4 = 2x - 6 \)
Add 6: \( 10 = 2x \)
Divide by 2: \( x = 5 \)? Wait, no, wait the options for BC are 2.5,6,10? Wait maybe I miscalculated. Wait, wait the options for \( x \)? Wait the problem's dropdown for \( x \) and BC. Wait let's recheck:

Wait \( ED = x + 2 \), \( BA = 4x - 6 \). Midline theorem: \( ED=\frac{1}{2}BA \)
So \( x + 2=\frac{1}{2}(4x - 6) \)
\( x + 2 = 2x - 3 \) (Wait, \( \frac{1}{2}(4x - 6)=2x - 3 \), not 2x - 6. Oh! I made a mistake. Let's correct:

\( x + 2 = 2x - 3 \)
Subtract \( x \): \( 2 = x - 3 \)
Add 3: \( x = 5 \)? No, but the BC options are 2.5,6,10. Wait maybe the diagram has different markings. Wait, maybe \( BE = x \), \( EC = x \), so \( BC = 2x \). \( ED = x + 2 \), \( BA = 4x - 6 \). Midline theorem: \( ED=\frac{1}{2}BA \)
So \( x + 2=\frac{1}{2}(4x - 6) \)
Multiply both sides by 2: \( 2x + 4 = 4x - 6 \)
Subtract \( 2x \): \( 4 = 2x - 6 \)
Add 6: \( 10 = 2x \)
\( x = 5 \). Then \( BC = 2x = 10 \)? But the options for BC include 10. Wait, but maybe my initial midpoint was wrong. Wait, maybe D is midpoint, E is midpoint, so ED is midline, so \( ED=\frac{1}{2}AB \). So \( x + 2=\frac{1}{2}(4x - 6) \)
Solving: \( 2(x + 2)=4x - 6 \)
\( 2x + 4 = 4x - 6 \)
\( 4 + 6 = 4x - 2x \)
\( 10 = 2x \)
\( x = 5 \). Then \( BC = BE + EC = x + x = 2x = 10 \). So \( x = 5 \)? But the dropdown for \( x \) – maybe the options were different. Wait the problem's BC options are 2.5,6,10. So if \( x = 5 \), BC=10. Let's check again.

Wait maybe the equation was \( x + 2=\frac{1}{2}(4x - 6) \). Let's plug x=5: left side 5+2=7, right side (20-6)/2=7. Correct. Then BC=2x=10. So:

First blank: D (midpoint of AC)
Second blank: Midline (Midsegment)
Third blank: x=5 (but if the dropdown has 5? Wait the user's image shows BC options 2.5,6,10. So maybe I messed up. Wait maybe the BA is 4x - 6, ED is x + 2. Let's try x=5: BA=4*5 -6=14, ED=5+2=7, 7=14/2=7. Correct. Then BC=2x=10. So:

• Midpoint of AC: D
• Theorem: Midline (Midsegment)
• x=5 (but if the dropdown for x has 5, or maybe the problem had a typo. Wait the BC options are 2.5,6,10. So 2x=10 when x=5. So:

Answer:

First blank: D
Second blank: Midline (or Midsegment)
Third blank: \( x = 5 \) (if option)
Fourth blank: \( BC = 10 \) (since \( BC = 2x = 2*5 = 10 \))

But following the problem's dropdown for BC, the answer is 10. So final answer for BC is 10.