the diagram shows isosceles trapezoid lmnp. it also shows how line segment no was drawn to form parallelogram lmno. what is the measure of angle onp? 50° 65° 80° 130°

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# Explanation:
## Step1: Recall properties of parallelograms
In parallelogram $ LMNO $, $ \angle L = \angle MNO = 50^\circ $ (opposite angles of a parallelogram are equal) and $ LM \parallel ON $, $ LN \parallel MO $. Also, in isosceles trapezoid $ LMNP $, $ LM \parallel NP $, so $ ON \parallel NP $? Wait, no, $ LM \parallel NP $ (bases of isosceles trapezoid) and $ LM \parallel ON $ (sides of parallelogram), so $ ON \parallel LM \parallel NP $? Wait, actually, $ LMNP $ is isosceles trapezoid, so $ \angle L $ and $ \angle P $ are related? Wait, no, let's correct. In isosceles trapezoid $ LMNP $, $ LM \parallel NP $, and $ \angle L + \angle M = 180^\circ $, $ \angle L = \angle P $? Wait, no, in isosceles trapezoid, base angles are equal. So $ \angle L = \angle P $? Wait, no, $ \angle L $ and $ \angle M $ are adjacent angles on base $ LM $, so they are supplementary. Wait, maybe better to use the parallelogram first.

In parallelogram $ LMNO $, $ \angle L = 50^\circ $, so $ \angle MON $? Wait, no, $ \angle L = 50^\circ $, and $ LM \parallel ON $, so $ \angle L + \angle LON = 180^\circ $? No, $ LM \parallel ON $, so $ \angle L $ and $ \angle LON $ are same - side interior angles? Wait, no, $ L $ is a vertex, $ LM $ and $ LN $ are sides. Wait, maybe $ \triangle ONP $ is isosceles? Because $ ON = LM $ (opposite sides of parallelogram), and in isosceles trapezoid $ LMNP $, $ LM = PN $? Wait, no, in isosceles trapezoid, the non - parallel sides are equal, so $ LN = MP $. Wait, maybe $ ON = PN $, so $ \triangle ONP $ is isosceles with $ ON = PN $.

Since $ LMNO $ is a parallelogram, $ ON = LM $. In isosceles trapezoid $ LMNP $, $ LM = PN $ (wait, no, the legs are equal, but the bases: $ LM $ and $ NP $ are the bases? Wait, no, in trapezoid $ LMNP $, the bases are $ LM $ and $ NP $, and the legs are $ LN $ and $ MP $. Wait, maybe I made a mistake. Let's start over.

1. In parallelogram $ LMNO $: $ \angle L = 50^\circ $, so $ \angle MNO = 50^\circ $ (opposite angles of parallelogram). Also, $ LM \parallel ON $, so $ \angle L + \angle LON = 180^\circ $? No, $ LM \parallel ON $, and $ LN $ is a transversal? Wait, $ L $, $ O $, $ P $ are colinear? Yes, from the diagram, $ L - O - P $ is a straight line. So $ LO $ is a segment, $ OP $ is another segment, so $ LOP $ is a straight line, so $ \angle LON + \angle NOP = 180^\circ $, but maybe not. Wait, in isosceles trapezoid $ LMNP $, $ LM \parallel NP $, and $ LMNO $ is a parallelogram, so $ LM \parallel ON $, hence $ ON \parallel NP $? No, $ ON \parallel LM $ and $ LM \parallel NP $, so $ ON \parallel NP $, which would mean $ O $, $ N $, $ P $ are colinear, but that's not possible. Wait, no, $ LMNP $ is a trapezoid, so $ LM $ and $ NP $ are the two bases (parallel sides), and $ LN $ and $ MP $ are the legs. Then $ LMNO $ is a parallelogram, so $ LM \parallel ON $ and $ LM = ON $, so $ ON \parallel LM \parallel NP $, so $ ON $ is parallel to $ NP $, which would imply $ O $, $ N $, $ P $ are colinear, but that's not the case. Wait, the diagram shows $ L - O - P $ on the base, so $ LO $ and $ OP $ are parts of the base $ LP $. So $ LMNO $ has sides $ LM $, $ MN $, $ NO $, $ OL $. So $ LM \parallel NO $ and $ MN \parallel OL $. So $ \angle L = 50^\circ $, so $ \angle NO L = 50^\circ $? No, in parallelogram, opposite angles are equal, so $ \angle L = \angle MNO = 50^\circ $, and $ \angle OLM = \angle ONM $? Wait, maybe better to look at triangle $ ONP $.

Since $ LMNP $ is isosceles trapezoid, $ \angle L = \angle P = 50^\circ $? Wait, no, in isosceles trapezoid, base angles are equal. So if $ LM $ and $ NP $ are the bases, then $ \angle L $ and $ \angle M $ are adjacent to base $ LM $, and $ \angle N $ and $ \angle P $ are adjacent to base $ NP $. Wait, maybe I got the bases wrong. Let's assume $ LP $ and $ MN $ are the legs, and $ LM $ and $ NP $ are the bases. So $ LM \parallel NP $, and $ LP = MN $ (legs of isosceles trapezoid). Then in parallelogram $ LMNO $, $ LM = ON $ (opposite sides of parallelogram), so $ ON = LM $, and since $ LMNP $ is isosceles trapezoid, $ LM = NP $? No, that's not true for trapezoid. Wait, maybe $ \triangle ONP $ is isosceles with $ ON = PN $, so $ \angle ONP=\angle NOP $.

Wait, another approach: In isosceles trapezoid $ LMNP $, $ \angle L + \angle M = 180^\circ $, but $ LMNO $ is a parallelogram, so $ \angle M = \angle LON $ (opposite angles of parallelogram). Wait, $ \angle L = 50^\circ $, so $ \angle M = 180^\circ - 50^\circ = 130^\circ $ (since $ LM $ and $ MN $ are adjacent sides of trapezoid, supplementary). Then in parallelogram $ LMNO $, $ \angle M = \angle LON = 130^\circ $? No, opposite angles of parallelogram are equal, so $ \angle L = \angle MNO = 50^\circ $, $ \angle M = \angle LON = 130^\circ $. Now, $ LP $ is a straight line, so $ \angle LON + \angle NOP = 180^\circ $, so $ \angle NOP = 180^\circ - 130^\circ = 50^\circ $. Now, in $ \triangle ONP $, since $ LMNP $ is isosceles trapezoid, $ PN = LM $, and $ LM = ON $ (parallelogram), so $ PN = ON $, so $ \triangle ONP $ is isosceles with $ ON = PN $. Therefore, $ \angle ONP=\angle NOP $? Wait, no, in $ \triangle ONP $, angles sum to $ 180^\circ $. Wait, if $ ON = PN $, then $ \angle ONP $ and $ \angle NOP $ are the base angles? Wait, no, the sides $ ON $ and $ PN $ are equal, so the angles opposite them are equal. The angle opposite $ ON $ is $ \angle P $, and the angle opposite $ PN $ is $ \angle NOP $. Wait, I think I messed up.

Wait, let's use the fact that in isosceles trapezoid $ LMNP $, $ \angle L = \angle P = 50^\circ $ (base angles). Wait, no, base angles are the angles adjacent to each base. If $ LM $ and $ NP $ are the bases, then $ \angle L $ and $ \angle M $ are adjacent to base $ LM $, $ \angle N $ and $ \angle P $ are adjacent to base $ NP $, and $ \angle L = \angle N $, $ \angle M = \angle P $? No, that's not right. In isosceles trapezoid, base angles are equal. So $ \angle L = \angle P $ and $ \angle M = \angle N $. Wait, yes! In isosceles trapezoid, the base angles are equal. So $ \angle L = \angle P = 50^\circ $, and $ \angle M = \angle N = 130^\circ $. Now, $ LMNO $ is a parallelogram, so $ \angle L = \angle MNO = 50^\circ $. Then $ \angle ONP=\angle MNP - \angle MNO $. Since $ \angle MNP = 130^\circ $ (base angle of isosceles trapezoid), then $ \angle ONP = 130^\circ - 50^\circ = 80^\circ $? Wait, no, that doesn't make sense. Wait, maybe $ \angle MNP $ is $ 130^\circ $, and $ \angle MNO = 50^\circ $, so $ \angle ONP = 180^\circ - 50^\circ - 50^\circ = 80^\circ $? Wait, let's start over with correct properties.

1. Properties of parallelogram $ LMNO $:
   - $ LM \parallel ON $ and $ LN \parallel MO $.
   - $ \angle L=\angle MNO = 50^\circ $ (opposite angles of a parallelogram are equal).
2. Properties of isosceles trapezoid $ LMNP $:
   - $ LM \parallel NP $, so $ ON \parallel NP $ (since $ LM \parallel ON $ from parallelogram). Wait, no, $ LM \parallel NP $ (bases of trapezoid) and $ LM \parallel ON $ (sides of parallelogram), so $ ON \parallel NP $, which would mean $ O $, $ N $, $ P $ are colinear, but that's not possible. So actually, $ LMNP $ has $ LM $ and $ NP $ as the two parallel sides (bases), and $ LN $ and $ MP $ as the non - parallel sides (legs), with $ LN = MP $ (isosceles trapezoid). Also, $ \angle L+\angle M = 180^\circ $, $ \angle L=\angle P $, $ \angle M=\angle N $.
3. Since $ LMNO $ is a parallelogram, $ LM = ON $ and $ LM \parallel ON $. Since $ LMNP $ is an isosceles trapezoid, $ LM = NP $? No, in an isosceles trapezoid, the legs are equal, not the bases. Wait, the bases can be of different lengths. But $ ON = LM $ (from parallelogram) and in the trapezoid, $ \triangle ONP $ is formed by drawing $ NO $. Since $ LM \parallel NP $ and $ LM \parallel ON $, then $ ON $ is parallel to $ NP $, so $ \angle NOP=\angle P = 50^\circ $ (corresponding angles). And since $ ON = LM $ and in isosceles trapezoid $ LN = MP $, and $ LMNO $ is a parallelogram, $ LN = MO $, so $ MO = MP $, so $ \triangle MOP $ is isosceles? No, maybe $ ON = PN $, so $ \triangle ONP $ is isosceles with $ ON = PN $. Then, using the angle sum property of a triangle: $ \angle ONP + \angle NOP+\angle P = 180^\circ $. We know $ \angle NOP=\angle L = 50^\circ $ (corresponding angles, since $ ON \parallel LM $ and $ LP $ is a transversal) and $ \angle P=\angle L = 50^\circ $ (base angles of isosceles trapezoid). So $ \angle ONP=180^\circ - 50^\circ - 50^\circ = 80^\circ $.

# Answer:
$ 80^\circ $

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Explanation:

Step1: Recall properties of parallelograms

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