triangle opq is formed by connecting the midpoints of the side of triangle lmn. the measures of the interior angles of triangle lmn are shown. find the measure of ∠jqo. figures not necessarily drawn to scale.

Question

Accepted Answer

# Explanation:
## Step1: Find angle at M in triangle LMN
The sum of angles in a triangle is $180^\circ$. In $\triangle LMN$, we know $\angle L = 80^\circ$ and $\angle N = 46^\circ$. Let $\angle M=x$. Then $80 + 46+x=180$. Solving for $x$, we get $x = 180-(80 + 46)=180 - 126 = 54^\circ$ (which matches the given angle at M, so that's a check).

## Step2: Use midsegment theorem (parallel lines)
Since $O$, $P$, $Q$ are midpoints, $PQ\parallel LM$ and $OQ\parallel MN$ (by triangle midsegment theorem: the segment connecting midpoints of two sides is parallel to the third side). So, $\angle LQO$ and $\angle N$ are corresponding angles (because $OQ\parallel MN$ and $LN$ is a transversal). Wait, no, wait. Wait, $O$ is midpoint of $LM$, $Q$ is midpoint of $LN$, so $OQ$ is midsegment, so $OQ\parallel MN$. Therefore, $\angle LQO=\angle N$? Wait, no, let's re - examine. Wait, $\angle L = 80^\circ$, $\angle N = 46^\circ$, $\angle M = 54^\circ$. Since $Q$ is midpoint of $LN$ and $O$ is midpoint of $LM$, then $OQ\parallel MN$ (midsegment theorem: $OQ$ connects midpoints of $LN$ and $LM$, so it's parallel to $MN$). Therefore, $\angle LQO$ and $\angle MNL$ (angle at N) are corresponding angles? Wait, no, when $OQ\parallel MN$, the transversal is $LN$. So $\angle LQO$ and $\angle N$ (angle at N) are equal? Wait, no, $\angle LQO$ is at $Q$ on $LN$, between $L$ and $Q$, and $OQ$ is going to $O$ on $LM$. Wait, maybe another approach. Wait, the triangle $OPQ$ is formed by midpoints, so $OQ\parallel MN$, so $\angle LQO=\angle N$? Wait, no, $\angle N$ is $46^\circ$, but wait, $\angle L = 80^\circ$, $\angle M = 54^\circ$, $\angle N = 46^\circ$. Wait, actually, since $OQ$ is midsegment, $OQ\parallel MN$, so $\angle LOQ=\angle M = 54^\circ$? No, maybe I messed up. Wait, let's look at triangle $LQO$. $OQ\parallel MN$, so $\angle LQO=\angle N = 46^\circ$? No, that doesn't seem right. Wait, no, the midsegment $OQ$ is parallel to $MN$, so the corresponding angle for $\angle LQO$ with respect to $MN$ would be $\angle N$. Wait, maybe I made a mistake. Wait, the sum of angles in triangle $LMN$ is $180$, we have $\angle L = 80$, $\angle M = 54$, $\angle N = 46$. Now, $Q$ is midpoint of $LN$, $O$ is midpoint of $LM$, so $OQ\parallel MN$ (midsegment theorem: $OQ\parallel MN$ and $OQ=\frac{1}{2}MN$). Therefore, $\angle LQO$ and $\angle N$ are equal (corresponding angles, since $OQ\parallel MN$ and $LN$ is transversal). Wait, but $\angle N$ is $46^\circ$, but wait, the question is $\angle LQO$. Wait, no, maybe $\angle LQO$ is equal to $\angle N$? Wait, no, let's think again. If $OQ\parallel MN$, then $\angle LQO=\angle MNL$ (angle at N), which is $46^\circ$? But that seems incorrect. Wait, no, $\angle L = 80^\circ$, and in triangle $LQO$, if $OQ\parallel MN$, then $\angle LOQ=\angle M = 54^\circ$ (corresponding angles, since $OQ\parallel MN$ and $LM$ is transversal). Then in triangle $LQO$, the sum of angles is $180^\circ$. So $\angle L + \angle LOQ+\angle LQO = 180$. We know $\angle L = 80^\circ$, $\angle LOQ = 54^\circ$, so $\angle LQO=180-(80 + 54)=180 - 134 = 46^\circ$? Wait, no, that's the same as $\angle N$. Wait, but maybe the correct approach is: since $OQ$ is midsegment, $OQ\parallel MN$, so $\angle LQO=\angle N = 46^\circ$? Wait, no, I think I confused the angles. Wait, let's start over.

The triangle midsegment theorem states that the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half as long. So, $Q$ is the midpoint of $LN$, $O$ is the midpoint of $LM$, so $OQ\parallel MN$. Therefore, $OQ\parallel MN$, so the corresponding angles formed by transversal $LN$ are equal. So $\angle LQO$ (at $Q$ on $LN$, between $L$ and $Q$) and $\angle MNL$ (at $N$ on $LN$, between $M$ and $N$) are equal. Since $\angle MNL = 46^\circ$, then $\angle LQO = 46^\circ$? Wait, no, that can't be, because $\angle L = 80^\circ$, and if we look at triangle $LQO$, if $OQ\parallel MN$, then $\angle LOQ=\angle LMN = 54^\circ$ (corresponding angles, transversal $LM$). Then in triangle $LQO$, $\angle L + \angle LOQ+\angle LQO=180$. So $80 + 54+\angle LQO = 180$, so $\angle LQO=180 - 134 = 46^\circ$. Wait, but that's the same as angle at N. Alternatively, since $OQ\parallel MN$, $\angle LQO=\angle N = 46^\circ$. Wait, but let's check with the other angle. Wait, maybe the answer is $46^\circ$? Wait, no, wait, $\angle L = 80^\circ$, $\angle M = 54^\circ$, $\angle N = 46^\circ$. Since $OQ$ is midsegment, $OQ\parallel MN$, so $\angle LQO=\angle N = 46^\circ$. Wait, but I think I made a mistake earlier. Wait, the midsegment $OQ$ is parallel to $MN$, so the angle $\angle LQO$ (at $Q$) is equal to the angle at $N$ (angle $N$) because they are corresponding angles (since $OQ\parallel MN$ and $LN$ is the transversal). So $\angle LQO = 46^\circ$? Wait, no, that seems wrong. Wait, no, $\angle L$ is $80^\circ$, $\angle M$ is $54^\circ$, $\angle N$ is $46^\circ$. Let's think about the lines: $O$ is midpoint of $LM$, $Q$ is midpoint of $LN$, so $OQ$ is midsegment, so $OQ\parallel MN$. Therefore, the direction of $OQ$ is same as $MN$. So from point $Q$, $OQ$ goes towards $O$ (midpoint of $LM$), and $MN$ goes from $M$ to $N$. So the angle between $LQ$ and $OQ$ (which is $\angle LQO$) should be equal to the angle between $LN$ and $MN$ (which is $\angle N$). So $\angle LQO=\angle N = 46^\circ$. Wait, but let's check with the triangle angle sum. In triangle $LQO$, we have $\angle L = 80^\circ$, $\angle LQO = 46^\circ$, then $\angle LOQ=180-(80 + 46)=54^\circ$, which is equal to $\angle M = 54^\circ$, which makes sense because $OQ\parallel MN$ and $LM$ is transversal, so $\angle LOQ=\angle M$ (corresponding angles). So that checks out. So $\angle LQO = 46^\circ$? Wait, no, wait, $\angle LQO$ is at $Q$, between $L$ and $Q$, and $OQ$ is going to $O$ on $LM$. Wait, maybe I had the angle wrong. Wait, the question is $\angle LQO$, which is the angle at $Q$ between $L$ and $O$. Wait, no, the notation $\angle LQO$ means the angle at $Q$ with segments $LQ$ and $OQ$. So vertex at $Q$, sides $QL$ and $QO$. So since $OQ\parallel MN$, and $LN$ is the transversal, $\angle LQO$ (at $Q$, between $L$ and $Q$ on $LN$, and $QO$ going to $O$) is equal to $\angle N$ (at $N$, between $L$ and $N$ on $LN$, and $NM$ going to $M$) because they are corresponding angles (since $OQ\parallel MN$). So $\angle LQO=\angle N = 46^\circ$. Wait, but that seems to contradict my initial thought, but the angle sum in triangle $LQO$ gives $\angle LOQ = 54^\circ$, which is equal to $\angle M$, so that's consistent with $OQ\parallel MN$ (corresponding angles with transversal $LM$). So the measure of $\angle LQO$ is $46^\circ$? Wait, no, wait, $\angle N$ is $46^\circ$, $\angle M$ is $54^\circ$, $\angle L$ is $80^\circ$. Wait, maybe I mixed up the angles. Wait, let's do it again.

1. In $\triangle LMN$, $\angle L=80^\circ$, $\angle M = 54^\circ$, $\angle N=46^\circ$ (since $80 + 54+46 = 180$).
2. $Q$ is the midpoint of $LN$, $O$ is the midpoint of $LM$. By the midsegment theorem, $OQ\parallel MN$.
3. When two parallel lines ($OQ$ and $MN$) are cut by a transversal ($LN$), corresponding angles are equal. So $\angle LQO$ (corresponding to $\angle N$) is equal to $\angle N$.
4. Therefore, $\angle LQO=\angle N = 46^\circ$. Wait, but that seems too simple. Wait, no, maybe the angle is $\angle LQO=\angle M$? No, because the transversal for $\angle M$ would be $LM$. Let's take transversal $LM$: $OQ\parallel MN$, so $\angle LOQ$ (at $O$, between $L$ and $O$ on $LM$, and $OQ$ going to $Q$) and $\angle M$ (at $M$, between $L$ and $M$ on $LM$, and $MN$ going to $N$) are corresponding angles, so $\angle LOQ=\angle M = 54^\circ$. Then in $\triangle LQO$, $\angle L + \angle LOQ+\angle LQO=180$, so $80+54+\angle LQO = 180$, so $\angle LQO=180 - 134 = 46^\circ$. So that's consistent. So the measure of $\angle LQO$ is $46^\circ$.

# Answer:
$46^\circ$

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QUESTION IMAGE

Question

Explanation:

Step1: Find angle at M in triangle LMN

Step2: Use midsegment theorem (parallel lines)

Answer: