homework for triangle congruence by hl
why are the two triangles congruent? name the congruent triangles.
1)
2) b is the midpoint of ad and ec
what additional information do you need to prove the triangles are congruent by hl?
3) δxrv ≅ δtrv
4) δsqr ≅ δmqr
5) δtsf ≅ δstr

Question

Accepted Answer

### Problem 1:
# Explanation:
## Step1: Identify Right Triangles
Both $\triangle PRS$ and $\triangle PRQ$ are right triangles (right angles at $R$ and $P$? Wait, looking at the diagram: $\angle R$ and $\angle P$ are right angles? Wait, the diagram has $S$, $R$ with right angle at $R$? Wait, no, the first diagram: $S$, $R$ (right angle at $R$), $P$ (right angle at $P$), $Q$. And sides $PS$ and $PQ$ are marked equal? Wait, the marks: $PS$ and $PQ$? Wait, the congruent sides: $SR$ and $QR$? Wait, no, the diagram shows $PS$ and $PQ$ with marks? Wait, maybe $\triangle PRS$ and $\triangle PRQ$: right angles at $R$ and $P$? Wait, no, the right angles are at $R$ (for $\triangle PRS$) and $P$ (for $\triangle PRQ$). Then hypotenuse $PR$ is common? Wait, no, $PR$ is a leg? Wait, maybe $\triangle SRP$ and $\triangle QRP$: right angles at $R$ and $P$, hypotenuse $SP = QP$ (marked), and leg $PR$ is common. So by HL (Hypotenuse-Leg), since they are right triangles, hypotenuse $SP = QP$ and leg $PR$ is common. So $\triangle SRP \cong \triangle QRP$ by HL.

## Step2: Name Congruent Triangles
From the diagram, the right triangles are $\triangle SRP$ and $\triangle QRP$ (or $\triangle PRS$ and $\triangle PRQ$). So the congruent triangles are $\triangle PRS \cong \triangle PRQ$ (or $\triangle SRP \cong \triangle QRP$) by HL (Hypotenuse-Leg) because they are right triangles, share the leg $PR$, and have equal hypotenuses $PS = PQ$ (marked).

# Answer:
The two triangles are congruent by HL (Hypotenuse - Leg) because they are right triangles, share the leg $ PR $, and have equal hypotenuses ($ PS = PQ $ as marked). The congruent triangles are $ \triangle PRS \cong \triangle PRQ $ (or $ \triangle SRP \cong \triangle QRP $).

### Problem 2:
# Explanation:
## Step1: Identify Midpoint Information
$B$ is the midpoint of $AD$ and $EC$, so $AB = BD$ and $EB = BC$. $\triangle AEB$ and $\triangle DCB$ are right triangles (right angles at $E$ and $C$). So we have $AB = BD$ (from midpoint), $EB = BC$ (from midpoint), and right angles. Wait, by HL? Wait, no, HL is for right triangles: hypotenuse and leg. Wait, here, $AB$ and $BD$ are legs? Wait, $\triangle AEB$ (right at $E$) and $\triangle DCB$ (right at $C$). So $EB = BC$ (leg), and $AB = BD$ (leg)? No, HL needs hypotenuse and leg. Wait, maybe $AE = DC$? No, wait, midpoint: $AB = BD$ and $EB = BC$. Then by SAS? Wait, no, the question is why congruent by HL. Wait, maybe $\triangle AEB$ and $\triangle DCB$: right angles, $EB = BC$ (leg), and $AB = BD$ (leg)? No, HL is hypotenuse and leg. Wait, maybe $AB$ and $BD$ are hypotenuses? Wait, no, $AB$ is a leg? Wait, the right angles are at $E$ and $C$, so the hypotenuses are $AB$ and $BD$? Wait, $AB$ is the hypotenuse of $\triangle AEB$ (right at $E$), and $BD$ is the hypotenuse of $\triangle DCB$ (right at $C$). Since $B$ is the midpoint, $AB = BD$ and $EB = BC$ (leg). So by HL, hypotenuse $AB = BD$ and leg $EB = BC$, so $\triangle AEB \cong \triangle DCB$ by HL.

## Step2: Name Congruent Triangles
The congruent triangles are $\triangle AEB \cong \triangle DCB$ by HL because they are right triangles, $EB = BC$ (leg), and $AB = BD$ (hypotenuse, since $B$ is midpoint of $AD$).

# Answer:
The two triangles are congruent by HL because they are right triangles, $ EB = BC $ (leg, as $ B $ is midpoint of $ EC $) and $ AB = BD $ (hypotenuse, as $ B $ is midpoint of $ AD $). The congruent triangles are $ \triangle AEB \cong \triangle DCB $.

### Problem 3:
# Explanation:
## Step1: HL Requirement
For $\triangle XRV \cong \triangle TRV$ by HL, we need a right triangle (they are right triangles, right angles at $X$ and $T$), a leg, and hypotenuse. We know $RV$ is common (hypotenuse? Or leg?). Wait, $\triangle XRV$ and $\triangle TRV$: right angles at $X$ and $T$, leg $RV$ is common? No, $RV$ is the hypotenuse? Wait, $RX$ and $RT$: no, $XV$ and $TV$? Wait, the right angles are at $X$ and $T$, so the legs are $RX, XV$ and $RT, TV$. The hypotenuse is $RV$ for both. Wait, but we need either $RX = RT$ (legs) or $XV = TV$ (legs), but since it's HL, we need hypotenuse (which is $RV$, common) and a leg. Wait, but $RV$ is the hypotenuse, so we need one leg to be equal. Wait, the triangles are $\triangle XRV$ (right at $X$) and $\triangle TRV$ (right at $T$). So hypotenuse $RV$ is common. To prove by HL, we need one leg equal: either $XR = TR$ or $XV = TV$. But looking at the diagram, $RV$ is the common hypotenuse, so we need $XR = TR$ (the other legs) or $XV = TV$. Wait, but the problem is "What additional information do you need to prove the triangles are congruent by HL?" So since they are right triangles, hypotenuse $RV$ is common, so we need one leg equal: either $ XR = TR $ or $ XV = TV $. But from the diagram, $XR$ and $TR$ are the legs adjacent to the right angles. So the additional information needed is $ XR = TR $ (or $ XV = TV $). But since it's HL, hypotenuse $RV$ is common, so we need a leg: $ XR = TR $.

## Step2: Conclusion
To prove $\triangle XRV \cong \triangle TRV$ by HL, we need the leg $ XR = TR $ (since they are right triangles with hypotenuse $ RV $ common, HL requires hypotenuse and one leg; hypotenuse is common, so we need one leg equal, e.g., $ XR = TR $).

# Answer:
To prove $ \triangle XRV \cong \triangle TRV $ by HL, we need the additional information that $ XR = TR $ (or $ XV = TV $; since they are right triangles with hypotenuse $ RV $ common, HL requires hypotenuse and one leg, so a leg like $ XR = TR $ is needed).

### Problem 4:
# Explanation:
## Step1: Identify Right Triangles
$\triangle SQR$ and $\triangle MQR$ are right triangles (right angle at $Q$ for both, since $RQ \perp SM$). So they are right triangles with right angle at $Q$.

## Step2: HL Requirement
For HL, we need hypotenuse and one leg equal. The hypotenuse would be $SR$ and $MR$, or the legs $SQ$ and $MQ$, $RQ$ (common). Wait, $RQ$ is a common leg. So to prove by HL, we need either hypotenuse $SR = MR$ or leg $SQ = MQ$. But the problem is "What additional information do you need to prove the triangles are congruent by HL?" So since they are right triangles (right at $Q$), common leg $RQ$, so we need hypotenuse $SR = MR$ or leg $SQ = MQ$. But HL requires hypotenuse and leg, so if we have $RQ$ (leg) common, we need hypotenuse $SR = MR$, or if we have hypotenuse $SR = MR$, then leg $RQ$ is common. Wait, no: HL is hypotenuse and one leg. So for $\triangle SQR$ (right at $Q$) and $\triangle MQR$ (right at $Q$), legs: $SQ$, $RQ$ and $MQ$, $RQ$. Hypotenuse: $SR$ and $MR$. So to prove by HL, we need either $SR = MR$ (hypotenuse) and $RQ$ (leg) common, or $SQ = MQ$ (leg) and $RQ$ (leg) common? No, HL is hypotenuse and one leg. So if we take hypotenuse $SR = MR$ and leg $RQ$ (common), then HL applies. Or leg $SQ = MQ$ and hypotenuse $SR = MR$? Wait, no, HL is hypotenuse and a leg, not two legs (that would be SAS). So the additional information needed is $ SR = MR $ (hypotenuse) or $ SQ = MQ $ (leg). But since $RQ$ is common, we need either hypotenuse $ SR = MR $ or leg $ SQ = MQ $. But the problem is about HL, so hypotenuse and leg. So the additional information is $ SR = MR $ (hypotenuse) or $ SQ = MQ $ (leg). But looking at the diagram, $RQ$ is perpendicular to $SM$, so $RQ$ is a leg. So to use HL, we need hypotenuse $SR = MR$ (since $RQ$ is common leg).

## Step2: Conclusion
To prove $\triangle SQR \cong \triangle MQR$ by HL, we need the additional information that $ SR = MR $ (hypotenuse) or $ SQ = MQ $ (leg). Since $RQ$ is the common leg, the hypotenuse $ SR = MR $ would satisfy HL.

# Answer:
To prove $ \triangle SQR \cong \triangle MQR $ by HL, we need the additional information that $ SR = MR $ (or $ SQ = MQ $; since they are right triangles with leg $ RQ $ common, HL requires hypotenuse and one leg, so hypotenuse $ SR = MR $ or leg $ SQ = MQ $ is needed).

### Problem 5:
# Explanation:
## Step1: Identify Right Triangles
$\triangle TSF$ (right at $S$) and $\triangle STR$ (right at $T$)? Wait, the diagram: $S$ has right angle, $T$ has right angle, $F$, $S$, $T$, $R$. So $\triangle TSF$ (right at $S$) and $\triangle STR$ (right at $T$). Wait, no, the right angles are at $S$ (for $\triangle TSF$) and $T$ (for $\triangle STR$). The common side is $ST$. So hypotenuse would be $TF$ and $SR$? Wait, no, the triangles are $\triangle TSF$ (right at $S$: legs $TS$, $SF$; hypotenuse $TF$) and $\triangle STR$ (right at $T$: legs $ST$, $TR$; hypotenuse $SR$). Wait, no, maybe $\triangle FST$ (right at $S$) and $\triangle RTS$ (right at $T$). The common side is $ST$. To prove by HL, we need hypotenuse and leg. So either $TF = SR$ (hypotenuse) and $ST$ (leg) common, or $SF = TR$ (leg) and $ST$ (leg) common. But HL requires hypotenuse and one leg. So the additional information needed is either $ TF = SR $ (hypotenuse) or $ SF = TR $ (leg).

## Step2: Conclusion
To prove $\triangle TSF \cong \triangle STR$ by HL, we need the additional information that $ TF = SR $ (hypotenuse) or $ SF = TR $ (leg). Since $ST$ is the common leg, we need either the hypotenuse $ TF = SR $ or the other leg $ SF = TR $.

# Answer:
To prove $ \triangle TSF \cong \triangle STR $ by HL, we need the additional information that $ TF = SR $ (or $ SF = TR $; since they are right triangles with leg $ ST $ common, HL requires hypotenuse and one leg, so hypotenuse $ TF = SR $ or leg $ SF = TR $ is needed).

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QUESTION IMAGE

Question

Explanation:

Problem 1:

Step1: Identify Right Triangles

Step2: Name Congruent Triangles

Step1: Identify Midpoint Information

Step2: Name Congruent Triangles

Step1: HL Requirement

Step2: Conclusion

Answer:

Problem 2: