eden is cutting two triangular tiles for her bathroom. she needs the tiles to be congruent but is not sure she is cutting them that way. eden has ensured that one side of both tiles is congruent. which pair of sides would eden need to compare in order to make sure the triangles are congruent by hl? image of a grid with two right triangles labeled a, b, c, d, e, f (with lengths 3.2) follows options: (overline{ac}) and (overline{fd}), (overline{ab}) and (overline{ef}), (overline{bc}) and (overline{ef}), (overline{bc}) and (overline{fd})

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# Explanation:
## Step1: Recall HL Congruence
HL (Hypotenuse - Leg) congruence applies to right - triangles. For two right - triangles to be congruent by HL, the hypotenuse and one leg of one right - triangle must be congruent to the hypotenuse and the corresponding leg of the other right - triangle.
First, identify the right - triangles. Triangle $ABC$ and triangle $DEF$ (assuming the right - triangles in the grid) are right - triangles with right angles at $A$ and $F$ respectively (from the diagram: $ \angle A = 90^{\circ}$ and $ \angle F=90^{\circ}$).
The hypotenuse of a right - triangle is the side opposite the right angle. In triangle $ABC$, the hypotenuse is $BC$ (length 3.2 as given), and in triangle $DEF$, the hypotenuse is $DE$? Wait, no, looking at the options, let's re - examine. Wait, the given sides with length 3.2: $BC = 3.2$ and the other triangle has a side of length 3.2 (probably the hypotenuse or leg).
Wait, let's analyze each option:
- Option 1: $\overline{AC}$ and $\overline{FD}$: $\overline{AC}$ is a leg of triangle $ABC$, $\overline{FD}$ is a leg of triangle $DEF$, but we need hypotenuse and leg.
- Option 2: $\overline{AB}$ and $\overline{EF}$: $\overline{AB}$ is a leg of triangle $ABC$, $\overline{EF}$ is a leg of triangle $DEF$, not hypotenuse - leg.
- Option 3: $\overline{BC}$ and $\overline{EF}$: $\overline{BC}$ is the hypotenuse of triangle $ABC$ (since $ \angle A = 90^{\circ}$), $\overline{EF}$ is a leg of triangle $DEF$ (since $ \angle F = 90^{\circ}$)? Wait, no, wait the right - angle in triangle $DEF$ is at $F$, so the legs are $FE$ and $FD$, hypotenuse is $DE$? Wait, maybe I misidentified. Wait, looking at the diagram, triangle $ABC$: right angle at $A$, so legs $AB$ and $AC$, hypotenuse $BC$. Triangle $DEF$: right angle at $F$, so legs $FE$ and $FD$, hypotenuse $DE$? But the length 3.2 is given for $BC$ and another side. Wait, the options: $\overline{BC}$ (hypotenuse of $ABC$) and $\overline{EF}$ (leg of $DEF$)? No, wait $\overline{EF}$ is a leg, $\overline{BC}$ is hypotenuse. Wait, no, let's check the other option: $\overline{BC}$ and $\overline{FD}$: $\overline{BC}$ is hypotenuse of $ABC$, $\overline{FD}$ is leg of $DEF$? No, wait maybe the right - triangle in the lower part: right angle at $F$, so legs are $FE$ (horizontal) and $FD$ (vertical), hypotenuse $DE$. The upper triangle: right angle at $A$, legs $AB$ (horizontal) and $AC$ (vertical), hypotenuse $BC$ (length 3.2). The lower triangle has a side of length 3.2 (probably the hypotenuse? Wait, no, the length 3.2 is marked on $BC$ and another side. Wait, maybe the two triangles: triangle $ABC$ (right - angled at $A$) and triangle $DFE$ (right - angled at $F$).
HL requires hypotenuse and one leg. So for triangle $ABC$ (right at $A$) and triangle $DFE$ (right at $F$):
- Hypotenuse of $ABC$ is $BC$, hypotenuse of $DFE$ would be $DE$, but $BC = 3.2$, maybe the other triangle's hypotenuse is also 3.2? Wait, the options: $\overline{BC}$ (hypotenuse of $ABC$) and $\overline{EF}$: $\overline{EF}$ is a leg. Wait, no, let's re - evaluate the options:
Wait, the correct HL pair: in right - triangles, if we have two right - triangles, and we know one leg is congruent (given), then to use HL, we need to compare the hypotenuses and the other leg? No, HL is hypotenuse and one leg. So if one leg is already congruent (given), then we need to compare the hypotenuses and the other leg? Wait, no, the problem says "Eden has ensured that one side of both tiles is congruent". So we need to find the other pair (hypotenuse and leg) for HL.
Looking at the diagram:
- Triangle $ABC$: right angle at $A$, so $AB$ and $AC$ are legs, $BC$ is hypotenuse.
- Triangle $DFE$: right angle at $F$, so $FE$ and $FD$ are legs, $DE$ is hypotenuse.
Wait, the length of $BC$ is 3.2, and the length of $DE$ (if it's the hypotenuse) - no, the other triangle has a side of length 3.2. Wait, the options:
- $\overline{AB}$ and $\overline{EF}$: $AB$ is a leg of $ABC$, $EF$ is a leg of $DFE$ (since $F$ is right angle, $EF$ is horizontal leg, $AB$ is horizontal leg of $ABC$). But HL needs hypotenuse and leg.
- $\overline{BC}$ and $\overline{EF}$: $BC$ is hypotenuse of $ABC$, $EF$ is leg of $DFE$ - no.
- $\overline{BC}$ and $\overline{FD}$: $BC$ is hypotenuse of $ABC$, $FD$ is leg of $DFE$ - no.
Wait, maybe I made a mistake. Wait, the right - triangle in the upper part: $A(1,4)$, $B(2,4)$, $C(1,1)$ (so $AB$ is from $(1,4)$ to $(2,4)$, length 1; $AC$ is from $(1,4)$ to $(1,1)$, length 3; $BC$ is from $(2,4)$ to $(1,1)$, length $\sqrt{(2 - 1)^2+(4 - 1)^2}=\sqrt{1 + 9}=\sqrt{10}\approx3.2$, which matches the 3.2 given. The lower triangle: $F(1, - 2)$, $E(2, - 2)$, $D(2,1)$ (so $FE$ is from $(1, - 2)$ to $(2, - 2)$, length 1; $FD$ is from $(1, - 2)$ to $(2,1)$, length $\sqrt{(2 - 1)^2+(1+2)^2}=\sqrt{1 + 9}=\sqrt{10}\approx3.2$? Wait, no, $D$ is at $(2,1)$, $F$ is at $(1, - 2)$, so $FD$ length is $\sqrt{(2 - 1)^2+(1+2)^2}=\sqrt{1 + 9}=\sqrt{10}\approx3.2$, and $DE$ is from $(2,1)$ to $(2, - 2)$, length 3. Wait, the right - angle in the lower triangle is at $F$? No, the right - angle is at $F$ (since $FE$ is horizontal and $FD$ is vertical? No, $FE$ is horizontal ($y=-2$, $x$ from 1 to 2), $FD$ is from $(1, - 2)$ to $(2,1)$, which is not vertical. Wait, the right - angle in the lower triangle is at $E$? Because $FE$ is horizontal and $DE$ is vertical (from $(2,1)$ to $(2, - 2)$). So triangle $DEF$ has right angle at $E$, so legs $DE$ (vertical, length 3) and $EF$ (horizontal, length 1), hypotenuse $DF$ (length $\sqrt{1^2 + 3^2}=\sqrt{10}\approx3.2$). The upper triangle $ABC$ has right angle at $A$, legs $AB$ (horizontal, length 1) and $AC$ (vertical, length 3), hypotenuse $BC$ (length $\sqrt{1^2+3^2}=\sqrt{10}\approx3.2$).
So now, for HL congruence (Hypotenuse - Leg) for two right - triangles:
- Triangle $ABC$ (right at $A$): hypotenuse $BC$, leg $AB$ (or $AC$)
- Triangle $DEF$ (right at $E$): hypotenuse $DF$, leg $EF$ (or $DE$)

We know that one side is congruent. Let's check the options:
- Option $\overline{AB}$ and $\overline{EF}$: $AB$ is a leg of $ABC$ (length 1), $EF$ is a leg of $DEF$ (length 1) - but HL needs hypotenuse and leg. Wait, no, if we already have one leg congruent (say $AB\cong EF$), then we need to check the hypotenuse and the other leg? No, HL is hypotenuse and one leg. Wait, maybe the given congruent side is a leg, and we need to check the hypotenuse and the other leg? No, the problem says "which pair of sides would Eden need to compare in order to make sure the triangles are congruent by HL".
HL states that if the hypotenuse and one leg of a right - triangle are congruent to the hypotenuse and one leg of another right - triangle, then the triangles are congruent.
In triangle $ABC$ (right at $A$) and triangle $DEF$ (right at $E$):
- Hypotenuse of $ABC$ is $BC$, hypotenuse of $DEF$ is $DF$ (both length $\sqrt{10}\approx3.2$)
- Leg of $ABC$: $AB$ (length 1), leg of $DEF$: $EF$ (length 1)
- Leg of $ABC$: $AC$ (length 3), leg of $DEF$: $DE$ (length 3)

Now, looking at the options:
- $\overline{AB}$ and $\overline{EF}$: these are the legs (length 1 each), and if we already know one side is congruent (maybe a leg), then to use HL, we need hypotenuse and leg. Wait, no, maybe the given congruent side is the hypotenuse? Wait, the length 3.2 is the hypotenuse ( $BC$ and $DF$). So if the hypotenuse is congruent ( $BC\cong DF$), then we need to check a leg. But the options:
Wait, the options are:
1. $\overline{AC}$ and $\overline{FD}$
2. $\overline{AB}$ and $\overline{EF}$
3. $\overline{BC}$ and $\overline{EF}$
4. $\overline{BC}$ and $\overline{FD}$

Wait, let's re - express the coordinates:
- $A(1,4)$, $B(2,4)$, $C(1,1)$
- $F(1, - 2)$, $E(2, - 2)$, $D(2,1)$

So:
- $\overline{AB}$: from $(1,4)$ to $(2,4)$, length $2 - 1=1$
- $\overline{EF}$: from $(1, - 2)$ to $(2, - 2)$, length $2 - 1 = 1$
- $\overline{AC}$: from $(1,4)$ to $(1,1)$, length $4 - 1=3$
- $\overline{FD}$: from $(1, - 2)$ to $(2,1)$, length $\sqrt{(2 - 1)^2+(1 + 2)^2}=\sqrt{1 + 9}=\sqrt{10}\approx3.2$
- $\overline{BC}$: from $(2,4)$ to $(1,1)$, length $\sqrt{(2 - 1)^2+(4 - 1)^2}=\sqrt{1 + 9}=\sqrt{10}\approx3.2$
- $\overline{DE}$: from $(2,1)$ to $(2, - 2)$, length $1-(-2)=3$

Now, HL: right - triangles, hypotenuse and one leg.
Triangle $ABC$ (right at $A$): hypotenuse $BC$ (length $\sqrt{10}$), leg $AB$ (length 1) or $AC$ (length 3)
Triangle $DEF$ (right at $E$): hypotenuse $DF$ (length $\sqrt{10}$), leg $EF$ (length 1) or $DE$ (length 3)

If we want to use HL, and we know that one side is congruent (say the hypotenuse $BC\cong DF$ since both are $\sqrt{10}$), then we need to check a leg. But the options: $\overline{AB}$ and $\overline{EF}$ are both length 1 (legs), so if $AB\cong EF$ and $BC\cong DF$ (hypotenuse), then by HL, the triangles are congruent. Wait, but the problem says "Eden has ensured that one side of both tiles is congruent". So maybe the one side is a leg, and we need to check the hypotenuse and the other leg? No, HL is hypotenuse and one leg. So if one leg is congruent (e.g., $AB\cong EF$), then we need to check the hypotenuse ($BC\cong DF$)? But $DF$ is not an option, but $BC$ is an option, and $EF$ is an option? Wait, no, $\overline{BC}$ and $\overline{EF}$: $BC$ is hypotenuse, $EF$ is leg - that's not HL. $\overline{AB}$ and $\overline{EF}$: both legs, and if the hypotenuses are congruent ( $BC$ and $DF$), but $DF$ is $\overline{FD}$ (since $F$ to $D$ is $FD$). Wait, $\overline{BC}$ is hypotenuse (length $\sqrt{10}$), $\overline{FD}$ is hypotenuse (length $\sqrt{10}$)? No, $FD$ is from $F(1, - 2)$ to $D(2,1)$, which is the hypotenuse of triangle $DFE$ (right at $E$), and $BC$ is the hypotenuse of triangle $ABC$ (right at $A$). So $BC\cong FD$ (both length $\sqrt{10}$), and if we have a leg congruent, say $AB\cong EF$ (both length 1), then by HL, the triangles are congruent. But the question is which pair to compare for HL. HL requires hypotenuse and one leg. So if we already have one leg congruent (given), then we need to compare the hypotenuses and the other leg? No, the problem is asking which pair to compare to use HL. So the correct pair should be hypotenuse and leg. Wait, in the options, $\overline{AB}$ and $\overline{EF}$ are legs (length 1), and $\overline{BC}$ and $\overline{FD}$ are hypotenuses (length $\sqrt{10}$). But HL is hypotenuse and one leg. So if we take hypotenuse $BC$ and leg $AB$ from triangle $ABC$, and hypotenuse $FD$ and leg $EF$ from triangle $DEF$, then if $BC\cong FD$ (hypotenuse) and $AB\cong EF$ (leg), then HL applies. But the options: $\overline{AB}$ and $\overline{EF}$ are legs, $\overline{BC}$ and $\overline{FD}$ are hypotenuses. Wait, maybe the right - triangles are $ABC$ (right at $A$) and $DFE$ (right at $F$)? No, $F$ is not a right angle. Wait, I think I made a mistake in identifying the right angle. Let's look at the diagram again: in triangle $ABC$, there is

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Step1: Recall HL Congruence

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Step1: Recall HL Congruence