in the figure below, points j, k, and l are the midpoints of the sides of $\triangle xyz$. suppose $yz = 76$, $jk = 12$, and $xy = 68$. find the following lengths. $jl = \square$ $xz = \square$ $lz = \square$

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Accepted Answer

# Explanation:
## Step1: Recall the Midline Theorem (Triangle Midsegment Theorem)
The Midline Theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. Also, since $ J, K, L $ are midpoints, we can use properties of midsegments.

### For $ JL $:
Since $ J $ is the midpoint of $ XY $ and $ L $ is the midpoint of $ XZ $, and $ K $ is the midpoint of $ YZ $? Wait, no, let's look at the sides. Wait, $ J $ is midpoint of $ XY $, $ K $ midpoint of $ YZ $, $ L $ midpoint of $ XZ $. Wait, the segment $ JK $: by midline theorem, $ JK $ should be parallel to $ XZ $ and $ JK=\frac{1}{2}XZ $? Wait, no, maybe $ JL $: since $ J $ is midpoint of $ XY $ and $ L $ is midpoint of $ XZ $, then $ JL $ is midline? Wait, no, let's re-examine.

Wait, the problem: $ J, K, L $ are midpoints of sides of $ \triangle XYZ $. So $ J $ is midpoint of $ XY $, $ K $ midpoint of $ YZ $, $ L $ midpoint of $ XZ $.

### For $ JL $:
Wait, $ YZ = 76 $, $ JK = 12 $, $ XY = 68 $.

Wait, $ JL $: since $ L $ is midpoint of $ XZ $ and $ J $ is midpoint of $ XY $, no, wait, maybe $ JL $ is equal to $ \frac{1}{2}YZ $? Wait, no. Wait, the midline theorem: the segment connecting midpoints of two sides is half the third side.

Wait, let's list the midpoints:

- $ J $: midpoint of $ XY $ (so $ XJ = JY = \frac{XY}{2} = \frac{68}{2} = 34 $)
- $ K $: midpoint of $ YZ $ (so $ YK = KZ = \frac{YZ}{2} = \frac{76}{2} = 38 $)
- $ L $: midpoint of $ XZ $ (so $ XL = LZ = \frac{XZ}{2} $)

Now, segment $ JK $: connects midpoints of $ XY $ and $ YZ $, so by midline theorem, $ JK \parallel XZ $ and $ JK = \frac{1}{2}XZ $. Wait, but $ JK = 12 $, so $ XZ = 2 \times JK = 2 \times 12 = 24 $? Wait, no, that can't be. Wait, maybe I got the sides wrong. Wait, maybe $ JK $ is midline for $ XZ $, so $ JK = \frac{1}{2}XZ $, so $ XZ = 2 \times 12 = 24 $. Then $ JL $: since $ J $ is midpoint of $ XY $ and $ L $ is midpoint of $ XZ $, then $ JL $ is midline for $ YZ $, so $ JL = \frac{1}{2}YZ $. Since $ YZ = 76 $, then $ JL = \frac{76}{2} = 38 $.

### Step1: Find $ JL $
Since $ J $ is midpoint of $ XY $ and $ L $ is midpoint of $ XZ $, by midline theorem, $ JL \parallel YZ $ and $ JL = \frac{1}{2}YZ $. Given $ YZ = 76 $, so $ JL = \frac{76}{2} = 38 $. Wait, no, wait, maybe $ JL $ is midline for $ YZ $. Wait, $ YZ $ is a side, so midline connecting midpoints of $ XY $ and $ XZ $ would be parallel to $ YZ $ and half its length. So $ JL = \frac{1}{2}YZ = \frac{76}{2} = 38 $.

### Step2: Find $ XZ $
Segment $ JK $ connects midpoints of $ XY $ and $ YZ $, so by midline theorem, $ JK = \frac{1}{2}XZ $. Given $ JK = 12 $, so $ XZ = 2 \times JK = 2 \times 12 = 24 $. Wait, that seems small, but let's check.

### Step3: Find $ LZ $
Since $ L $ is midpoint of $ XZ $, $ LZ = \frac{XZ}{2} $. We found $ XZ = 24 $, so $ LZ = \frac{24}{2} = 12 $. Wait, no, that contradicts? Wait, maybe I mixed up the midline.

Wait, let's re-express:

- Midline theorem: In a triangle, the segment connecting the midpoints of two sides is parallel to the third side and half as long.

So:

- $ JK $: connects midpoints of $ XY $ (J) and $ YZ $ (K), so third side is $ XZ $. Thus, $ JK = \frac{1}{2}XZ \implies XZ = 2 \times JK = 2 \times 12 = 24 $.

- $ JL $: connects midpoints of $ XY $ (J) and $ XZ $ (L), so third side is $ YZ $. Thus, $ JL = \frac{1}{2}YZ = \frac{76}{2} = 38 $.

- $ LZ $: since $ L $ is midpoint of $ XZ $, $ LZ = \frac{XZ}{2} = \frac{24}{2} = 12 $. Wait, but $ YZ = 76 $, $ XY = 68 $, $ XZ = 24 $. Let's check if the triangle is valid (triangle inequality: $ 68 + 24 > 76 $? $ 92 > 76 $, yes; $ 68 + 76 > 24 $, yes; $ 24 + 76 > 68 $, yes. So it's valid.

So:

- $ JL = 38 $ (since $ JL $ is midline of $ YZ $)
- $ XZ = 24 $ (since $ JK $ is midline of $ XZ $)
- $ LZ = 12 $ (since $ L $ is midpoint of $ XZ $)

Wait, but let's confirm again:

- $ J $ is midpoint of $ XY $: $ XJ = JY = 34 $
- $ K $ is midpoint of $ YZ $: $ YK = KZ = 38 $
- $ L $ is midpoint of $ XZ $: $ XL = LZ = 12 $ (since $ XZ = 24 $)
- $ JK $ connects $ J $ (mid $ XY $) and $ K $ (mid $ YZ $): by midline theorem, $ JK \parallel XZ $ and $ JK = \frac{1}{2}XZ $. So $ JK = 12 = \frac{1}{2} \times 24 $, which is correct.
- $ JL $ connects $ J $ (mid $ XY $) and $ L $ (mid $ XZ $): by midline theorem, $ JL \parallel YZ $ and $ JL = \frac{1}{2}YZ $. So $ JL = \frac{1}{2} \times 76 = 38 $, correct.
- $ LZ $ is half of $ XZ $, so $ 24 / 2 = 12 $, correct.

So the lengths are:

$ JL = 38 $

$ XZ = 24 $

$ LZ = 12 $

## Step1: Calculate $ JL $
By Midline Theorem, $ JL = \frac{1}{2}YZ $. Given $ YZ = 76 $, so $ JL = \frac{76}{2} = 38 $.

## Step2: Calculate $ XZ $
By Midline Theorem, $ JK = \frac{1}{2}XZ $. Given $ JK = 12 $, so $ XZ = 2 \times 12 = 24 $.

## Step3: Calculate $ LZ $
Since $ L $ is midpoint of $ XZ $, $ LZ = \frac{XZ}{2} $. Substituting $ XZ = 24 $, we get $ LZ = \frac{24}{2} = 12 $.

# Answer:
$ JL = \boldsymbol{38} $

$ XZ = \boldsymbol{24} $

$ LZ = \boldsymbol{12} $

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QUESTION IMAGE

Question

Explanation:

Step1: Recall the Midline Theorem (Triangle Midsegment Theorem)

For \( JL \):

For \( JL \):

Step1: Find \( JL \)

Step2: Find \( XZ \)

Step3: Find \( LZ \)

Answer: