explaining part of the proof of the triangle midsegment theorem
to prove part of the triangle midsegment theorem using the diagram, which statement must be shown?
○ the length of $overline{jk}$ equals the length of $overline{jl}$.
○ the length of $overline{gh}$ is half the length of $overline{kl}$.
○ the slope of $overline{jk}$ equals the slope of $overline{jl}$.
○ the slope of $overline{gh}$ is half the slope of $overline{kl}$.
(diagram: coordinate plane with points k(0, 0), l(2d, 0), j(2e, 2f), g(e, f), h(e + d, f))

Question

Accepted Answer

# Brief Explanations:
The Triangle Midsegment Theorem states that the midsegment (a segment connecting the midpoints of two sides) of a triangle is parallel to the third side and half its length. To prove part of this theorem using the diagram, we analyze the slopes (for parallelism, slopes must be equal) and lengths.

- For option 1: $ JK $ and $ JL $ are sides from the same vertex, not midsegments or related to the midsegment - third side relationship. Their lengths aren't necessarily equal.
- For option 2: $ GH $ is the midsegment, and $ KL $ is the base? Wait, no, $ KL $ is from $ K(0,0) $ to $ L(2d,0) $, length $ 2d $. $ GH $ is from $ G(e,f) $ to $ H(e + d,f) $, length $ d $. But the midsegment should be parallel to $ KL $ (the third side, which is $ KL $? Wait, the triangle is $ JKL $, with $ K(0,0) $, $ L(2d,0) $, $ J(2e,2f) $. The midpoints $ G $ (midpoint of $ JK $) and $ H $ (midpoint of $ JL $): midpoint of $ JK $: $ (\frac{0 + 2e}{2},\frac{0+2f}{2})=(e,f) $, midpoint of $ JL $: $ (\frac{2e+2d}{2},\frac{2f + 0}{2})=(e + d,f) $. So $ GH $ is the midsegment, and the third side is $ KL $. But the theorem says midsegment is parallel to third side (so slopes equal) and half its length. But to prove part of the theorem, we need to show parallelism (slope equality) or length relation. Wait, the slope of $ GH $: since $ G(e,f) $ and $ H(e + d,f) $, slope is $ \frac{f - f}{(e + d)-e}=0 $. Slope of $ KL $: $ K(0,0) $, $ L(2d,0) $, slope is $ \frac{0 - 0}{2d-0}=0 $. So slope of $ GH $ equals slope of $ KL $, but the options have slope of $ JK $ and $ JL $, or $ GH $ and $ KL $ slope. Wait, option 3: slope of $ JK $ and $ JL $. Slope of $ JK $: $ J(2e,2f) $, $ K(0,0) $, slope $ \frac{2f-0}{2e - 0}=\frac{f}{e} $. Slope of $ JL $: $ J(2e,2f) $, $ L(2d,0) $, slope $ \frac{2f-0}{2e-2d}=\frac{f}{e - d} $. These are not equal. Option 4: slope of $ GH $ is 0, slope of $ KL $ is 0, so slope of $ GH $ is equal (not half) of slope of $ KL $. Wait, maybe I misread the third side. Wait, the triangle is $ JKL $, with sides $ JK $, $ JL $, $ KL $. The midsegment $ GH $ connects midpoints of $ JK $ and $ JL $, so it should be parallel to $ KL $. So to prove part of the theorem (parallelism), we need to show that slope of $ GH $ equals slope of $ KL $, but the options: let's re - evaluate.

Wait, the options:

1. Length of $ JK $ equals length of $ JL $: $ JK $ length is $ \sqrt{(2e)^2+(2f)^2} $, $ JL $ length is $ \sqrt{(2d - 2e)^2+( - 2f)^2} $. Not necessarily equal.

2. Length of $ GH $ is half the length of $ KL $: $ GH $ length is $ d $ (since $ x $-coordinates differ by $ d $, $ y $-coordinates same), $ KL $ length is $ 2d $ ( $ x $-coordinates differ by $ 2d $, $ y $-coordinates same). So $ GH=\frac{1}{2}KL $. But is this the part to show? Wait, the midsegment theorem says midsegment is parallel and half the length. But to prove using the diagram, maybe the slope part? Wait, no, let's check the slopes again.

Slope of $ GH $: $ G(e,f) $, $ H(e + d,f) $, slope $ m_{GH}=\frac{f - f}{(e + d)-e}=0 $.

Slope of $ KL $: $ K(0,0) $, $ L(2d,0) $, slope $ m_{KL}=\frac{0 - 0}{2d-0}=0 $. So they are equal. But the options:

Option 3: slope of $ JK $ and $ JL $. Slope of $ JK $: $ \frac{2f-0}{2e - 0}=\frac{f}{e} $. Slope of $ JL $: $ \frac{2f-0}{2e-2d}=\frac{f}{e - d} $. Not equal.

Option 4: slope of $ GH $ is half the slope of $ KL $. But both slopes are 0, so 0 is not half of 0 (mathematically, 0 = 0, not half).

Wait, maybe the third side is $ KL $, and $ GH $ is midsegment. The length of $ GH $: distance between $ G(e,f) $ and $ H(e + d,f) $ is $ d $. Length of $ KL $: distance between $ K(0,0) $ and $ L(2d,0) $ is $ 2d $. So $ GH=\frac{1}{2}KL $. So option 2 says "The length of $ \overline{GH} $ is half the length of $ \overline{KL} $", which is correct. But wait, let's check the midsegment theorem again. The midsegment connects midpoints of two sides and is parallel to the third side and half its length. So to prove part of the theorem, showing that the length of the midsegment ($ GH $) is half the length of the third side ($ KL $) is a part of the proof.

Wait, but earlier slope calculation: $ GH $ and $ KL $ are horizontal lines (slope 0), so they are parallel. But the options:

Wait, maybe I made a mistake. Let's re - check the coordinates:

- $ K(0,0) $, $ L(2d,0) $: so $ KL $ is along the x - axis, length $ 2d $.

- $ J(2e,2f) $, $ G $ is midpoint of $ JK $: midpoint formula $ (\frac{x_1 + x_2}{2},\frac{y_1 + y_2}{2}) $, so $ G(\frac{0+2e}{2},\frac{0 + 2f}{2})=(e,f) $.

- $ H $ is midpoint of $ JL $: $ J(2e,2f) $, $ L(2d,0) $, so $ H(\frac{2e+2d}{2},\frac{2f + 0}{2})=(e + d,f) $.

So $ GH $ has coordinates from $ (e,f) $ to $ (e + d,f) $, so it's a horizontal line segment with length $ d $ (since $ \Delta x=d $, $ \Delta y = 0 $). $ KL $ has length $ 2d $ ( $ \Delta x = 2d $, $ \Delta y=0 $ ). So length of $ GH=\frac{1}{2}\times $ length of $ KL $. So option 2 is correct.

Wait, but let's check the other options again:

- Option 1: Length of $ JK $ and $ JL $: $ JK=\sqrt{(2e)^2+(2f)^2} $, $ JL=\sqrt{(2d - 2e)^2+( - 2f)^2} $. Not equal unless $ e = d - e $ (i.e., $ 2e=d $), which is not given.

- Option 3: Slope of $ JK $: $ \frac{2f-0}{2e-0}=\frac{f}{e} $, slope of $ JL $: $ \frac{2f - 0}{2e-2d}=\frac{f}{e - d} $. Not equal.

- Option 4: Slope of $ GH $ is 0, slope of $ KL $ is 0. So slope of $ GH $ is equal to slope of $ KL $, not half.

So the correct statement to show is that the length of $ \overline{GH} $ is half the length of $ \overline{KL} $.

# Answer:
The length of $\overline{GH}$ is half the length of $\overline{KL}$ (the second option).

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Answer: