determine whether each congruence statement is always true, sometimes true, or never true. move the answers to the correct boxes. (overline{lm} cong overline{mp}) (overline{jp} cong overline{qp}) (overline{hp} cong overline{hl}) (overline{jl} cong overline{jm}) (overline{mk} cong overline{mq}) always true sometimes true never true

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To solve this, we analyze each congruence statement based on the diagram (perpendicular lines, with $ M $ as the intersection, and points on the lines):

### 1. $ \boldsymbol{\overline{LM} \cong \overline{MP}} $
- $ M $ is the midpoint of $ \overline{LP} $ (since $ LP $ is a straight line through $ M $)? Wait, no—unless specified, $ LM $ and $ MP $ could be equal or not. But in the diagram, $ M $ is the intersection of two perpendicular lines, but $ LP $ is horizontal through $ M $. However, unless $ M $ is the midpoint, $ LM $ and $ MP $ might not be congruent. Wait, actually, in the diagram, $ M $ is the intersection, but $ LP $ is a straight line, so $ M $ is the midpoint? Wait, no—if $ LP $ is a straight line with $ M $ in the middle, then $ LM = MP $. Wait, maybe I misread. Wait, the diagram shows $ L $---$ M $---$ P $, so $ M $ is between $ L $ and $ P $. But is $ M $ the midpoint? The diagram has a right angle at $ M $ for the vertical line, but for $ LP $, unless marked, $ LM $ and $ MP $ could be equal or not. Wait, no—actually, in the diagram, $ LP $ is a straight line with $ M $ as the intersection, but $ LM $ and $ MP $ are segments from $ L $ to $ M $ and $ M $ to $ P $. If $ M $ is the midpoint, they are congruent; otherwise, not. But the problem is about "always true," "sometimes true," or "never true."

Wait, let's re-express:

- **$ \overline{LM} \cong \overline{MP} $**: $ M $ is on $ LP $, but unless $ M $ is the midpoint (not necessarily given), this is *sometimes true* (if $ M $ is the midpoint, it’s true; otherwise, false).

- **$ \overline{JP} \cong \overline{QP} $**: $ J $ and $ Q $ are on the vertical line through $ M $. $ JP $ is the distance from $ J $ to $ P $, $ QP $ from $ Q $ to $ P $. Since $ P $ is fixed, and $ J, Q $ are on the vertical line, $ JP $ and $ QP $ depend on $ J $ and $ Q $’s positions. Unless $ J $ and $ Q $ are equidistant from $ P $, this is *sometimes true*. Wait, no—wait, $ M $ is the intersection, so $ MP $ is horizontal, and $ J, Q $ are on vertical. $ JP = \sqrt{JM^2 + MP^2} $, $ QP = \sqrt{QM^2 + MP^2} $. So $ JP \cong QP $ only if $ JM = QM $, which is sometimes true (if $ J $ and $ Q $ are equidistant from $ M $).

- **$ \overline{HP} \cong \overline{HL} $**: $ H $ is on the vertical line, $ P $ and $ L $ are on horizontal. $ HP = \sqrt{HM^2 + MP^2} $, $ HL = \sqrt{HM^2 + LM^2} $. Since $ MP = LM $ (if $ M $ is midpoint of $ LP $), then $ HP = HL $. Wait, but earlier we thought $ LM $ and $ MP $ might be equal. Wait, maybe $ M $ is the midpoint of $ LP $, so $ LM = MP $. Then $ HP = \sqrt{HM^2 + LM^2} $, $ HL = \sqrt{HM^2 + LM^2} $, so $ HP \cong HL $ *always true*? Wait, no—$ H $ is on the vertical line, $ L $ and $ P $ are on horizontal, symmetric over $ M $. So $ HL $ and $ HP $ are both hypotenuses of right triangles with legs $ HM $ and $ LM $ (or $ MP $, since $ LM = MP $). So $ HL = HP $, so $ \overline{HP} \cong \overline{HL} $ is *always true*.

- **$ \overline{JL} \cong \overline{JM} $**: $ JL $ is the distance from $ J $ to $ L $ (right triangle: $ JM $ vertical, $ LM $ horizontal), so $ JL = \sqrt{JM^2 + LM^2} $, which is always longer than $ JM $ (since $ LM > 0 $). So $ JL \cong JM $ is *never true*.

- **$ \overline{MK} \cong \overline{MQ} $**: $ K $ and $ Q $ are on the vertical line through $ M $. $ MK $ is the distance from $ M $ to $ K $, $ MQ $ from $ M $ to $ Q $. Unless $ K $ and $ Q $ are equidistant from $ M $ (e.g., $ K $ above, $ Q $ below, but not necessarily), this is *sometimes true*. Wait, no—$ M $ is the intersection, $ K $ is above $ M $, $ Q $ is below $ M $. So $ MK $ and $ MQ $ are lengths from $ M $ to $ K $ (up) and $ M $ to $ Q $ (down). If $ K $ and $ Q $ are equidistant from $ M $, then $ MK \cong MQ $; otherwise, not. So *sometimes true*.

### Final Categorization:
- **Always True**: $ \overline{HP} \cong \overline{HL} $ (since $ L $ and $ P $ are symmetric over $ M $, so $ HL = HP $ by Pythagoras).
- **Sometimes True**: $ \overline{LM} \cong \overline{MP} $ (if $ M $ is midpoint), $ \overline{JP} \cong \overline{QP} $ (if $ J $ and $ Q $ are equidistant from $ P $), $ \overline{MK} \cong \overline{MQ} $ (if $ K $ and $ Q $ are equidistant from $ M $).
- **Never True**: $ \overline{JL} \cong \overline{JM} $ (since $ JL = \sqrt{JM^2 + LM^2} > JM $).

### Step-by-Step (for clarity):

#### 1. $ \boldsymbol{\overline{LM} \cong \overline{MP}} $
- Step 1: $ M $ lies on $ \overline{LP} $, but $ LM $ and $ MP $ are congruent only if $ M $ is the midpoint.
- Step 2: Since $ M $’s midpoint status is not guaranteed, this is sometimes true.

#### 2. $ \boldsymbol{\overline{JP} \cong \overline{QP}} $
- Step 1: $ JP = \sqrt{JM^2 + MP^2} $, $ QP = \sqrt{QM^2 + MP^2} $.
- Step 2: Congruent only if $ JM = QM $ (sometimes true).

#### 3. $ \boldsymbol{\overline{HP} \cong \overline{HL}} $
- Step 1: $ HL = \sqrt{HM^2 + LM^2} $, $ HP = \sqrt{HM^2 + MP^2} $.
- Step 2: Since $ LM = MP $ ( $ M $ is midpoint of $ LP $ ), $ HL = HP $ (always true).

#### 4. $ \boldsymbol{\overline{JL} \cong \overline{JM}} $
- Step 1: $ JL = \sqrt{JM^2 + LM^2} $, $ JM $ is a leg, $ JL $ is hypotenuse.
- Step 2: Hypotenuse > leg, so never congruent.

#### 5. $ \boldsymbol{\overline{MK} \cong \overline{MQ}} $
- Step 1: $ MK $ (up from $ M $) and $ MQ $ (down from $ M $) depend on $ K $ and $ Q $’s distance from $ M $.
- Step 2: Congruent only if $ K $ and $ Q $ are equidistant (sometimes true).

### Answer:
- **Always True**: $ \overline{HP} \cong \overline{HL} $
- **Sometimes True**: $ \overline{LM} \cong \overline{MP} $, $ \overline{JP} \cong \overline{QP} $, $ \overline{MK} \cong \overline{MQ} $
- **Never True**: $ \overline{JL} \cong \overline{JM} $

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QUESTION IMAGE

Question

Explanation:

1. \( \boldsymbol{\overline{LM} \cong \overline{MP}} \)

Answer: