Question

given δmno, find the measure of ∠lmn.

Explanation:

Step1: Identify the isosceles triangle

In $\triangle MNO$, $MN = ON$ (marked with equal segments), so it's isosceles with $\angle OMN=\angle ONM$. Also, $\angle PON = 104^\circ$ is a linear pair with $\angle NOM$, so $\angle NOM = 180^\circ - 104^\circ = 76^\circ$.

Step2: Calculate $\angle LMN$

In $\triangle MNO$, the sum of angles is $180^\circ$. Let $\angle OMN=\angle LMN = x$. Then $x + x + 76^\circ = 180^\circ$. Solving: $2x = 180^\circ - 76^\circ = 104^\circ$, so $x=\frac{104^\circ}{2}=52^\circ$? Wait, no—wait, $\angle LMN$ is $\angle OMN$? Wait, no, the diagram: $L - M - O - P$ is a straight line? Wait, maybe I messed up. Wait, $\angle POM$ is $104^\circ$, so $\angle NOM$ is supplementary? Wait, no, $L$, $M$, $O$, $P$ are colinear? So $\angle LMO$ is straight, so $\angle NMO$ and $\angle LMN$? Wait, no, $\triangle MNO$: $MN = ON$, so it's isosceles with base $MO$? Wait, no, the equal marks are on $MN$ and $ON$? Wait, the segments $MN$ and $ON$ are marked equal, so $\triangle MNO$ has $MN = ON$, so the base is $MO$, so the base angles are $\angle OMN$ and $\angle MON$? No, wait, in a triangle, equal sides have equal opposite angles. So if $MN = ON$, then the angles opposite them are $\angle MON$ and $\angle OMN$. Wait, $MN$ is opposite $\angle MON$, $ON$ is opposite $\angle OMN$. So $\angle MON=\angle OMN$. Wait, but $\angle PON$ is $104^\circ$, which is adjacent to $\angle MON$, so $\angle MON = 180^\circ - 104^\circ = 76^\circ$. Then $\angle OMN = 76^\circ$? No, that can't be. Wait, maybe the equal sides are $MN$ and $MO$? Wait, the diagram has two marks on $MN$ and $MO$? Wait, the original diagram: $N$ connected to $M$ and $O$, with $MN$ and $MO$ having equal marks? Wait, maybe I misread. Let's re-express:

Given $\triangle MNO$ is isosceles (two sides equal). The angle at $O$ adjacent to the straight line $L - M - O - P$: $\angle POM = 104^\circ$, so $\angle NOM = 180^\circ - 104^\circ = 76^\circ$. Now, if $\triangle MNO$ has $MN = MO$, then the base is $NO$, but no. Wait, the correct approach: in a straight line, $\angle LMN + \angle NMO = 180^\circ$? No, $L$, $M$, $O$, $P$ are colinear, so $LMOP$ is a straight line. So $\angle POM = 104^\circ$, so $\angle LMO = 180^\circ$, but $\angle NMO$ and $\angle NOM$: if $\triangle MNO$ is isosceles with $MN = ON$, then $\angle OMN = \angle NOM$? Wait, no, sides: $MN$ and $ON$ are equal, so angles opposite: $\angle MON$ (opposite $MN$) and $\angle OMN$ (opposite $ON$). So $\angle MON = \angle OMN$. Then $\angle MON = 180^\circ - 104^\circ = 76^\circ$, so $\angle OMN = 76^\circ$? But that's not one of the options. Wait, the options are 58, 52, 76, 104. Wait, maybe the equal sides are $MN$ and $MO$, so $\angle ONM = \angle NOM$. Then $\angle NOM = 180^\circ - 104^\circ = 76^\circ$, so $\angle ONM = 76^\circ$, then $\angle OMN = 180^\circ - 76^\circ - 76^\circ = 28^\circ$? No. Wait, maybe I got the straight line wrong. Wait, $L - M - O - P$: so $M$ is between $L$ and $O$, $O$ is between $M$ and $P$. So $\angle LMP$ is straight, so $\angle LMO$ is 180°, but $\angle OMP$ is 104°, so $\angle LMO$ is 180°, but $\angle LMN$ is part of $\angle LMO$. Wait, $\triangle MNO$: $MN = ON$, so it's isosceles with vertex at $N$. So base is $MO$, so $\angle NMO = \angle NOM$. Wait, $\angle NOM$ is supplementary to $\angle PON$ (104°), so $\angle NOM = 76°$, so $\angle NMO = 76°$, but then $\angle LMN$ would be 180° - 76° = 104°? No. Wait, maybe the equal sides are $MN$ and $MO$, so $\angle ONM = \angle NOM$. Then $\angle NOM = 180° - 104° = 76°$, so $\angle ONM = 76°$, then $\angle OMN = 180° - 76° - 76° = 28°$? Not an option. Wait, the options are 58, 52, 76, 104. Wait, maybe the angle at $O$ is 104°, so the exterior angle of $\triangle MNO$ at $O$ is 104°, so the remote interior angles sum to 104°. Since $\triangle MNO$ is isosceles, the two base angles are equal, so each is 104°/2 = 52°. Ah! That's it. The exterior angle theorem: the exterior angle is equal to the sum of the two non-adjacent interior angles. So $\angle PON$ is an exterior angle to $\triangle MNO$ at $O$, so $\angle PON = \angle LMN + \angle ONM$. But since $\triangle MNO$ is isosceles with $MN = ON$, $\angle LMN = \angle ONM$. So $104° = 2 \times \angle LMN$, so $\angle LMN = 104° / 2 = 52°$.

Step1: Apply Exterior Angle Theorem

In $\triangle MNO$, $\angle PON$ (104°) is an exterior angle. By the Exterior Angle Theorem, $\angle PON = \angle LMN + \angle ONM$.

Step2: Use Isosceles Triangle Property

Since $MN = ON$, $\triangle MNO$ is isosceles, so $\angle LMN = \angle ONM$ (let this angle be $x$).

Step3: Solve for $x$

Substitute into the exterior angle formula: $104^\circ = x + x$.
Simplify: $2x = 104^\circ$.
Divide: $x = \frac{104^\circ}{2} = 52^\circ$.

Answer:

$52^\circ$ (corresponding to the option with 52°)