Question

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°

Explanation:

Step1: Recall property of parallelogram

In parallelogram LMNO, $\angle L=\angle MNO = 50^{\circ}$ (opposite - angles of a parallelogram are equal).

Step2: Recall property of isosceles trapezoid

In isosceles trapezoid LMNP, $\angle L+\angle P=180^{\circ}$ (adjacent - base angles of an isosceles trapezoid are supplementary), so $\angle P = 180^{\circ}-\angle L=180 - 50=130^{\circ}$. Also, $\angle M+\angle N=180^{\circ}$ (adjacent - base angles of an isosceles trapezoid are supplementary).
In $\triangle NOP$, since LMNP is isosceles trapezoid and LMNO is parallelogram, $NO = LM$ and $LM = NP$ (properties of isosceles trapezoid and parallelogram), so $NO = NP$, and $\triangle NOP$ is isosceles.
$\angle NOP=\angle P = 130^{\circ}$ (base - angles of isosceles trapezoid).

Step3: Calculate $\angle ONP$

Using the angle - sum property of a triangle ($\angle NOP+\angle ONP+\angle NPO = 180^{\circ}$) in $\triangle NOP$. Let $\angle ONP=x$. Since $\angle NPO=\angle ONP$ (because $NO = NP$), we have $130^{\circ}+x + x=180^{\circ}$.
Combining like terms gives $130^{\circ}+2x=180^{\circ}$.
Subtract $130^{\circ}$ from both sides: $2x=180^{\circ}-130^{\circ}=50^{\circ}$.
Divide both sides by 2: $x = 25^{\circ}$. But we made a wrong start above.
Since LMNO is a parallelogram, $\angle L=\angle MNO = 50^{\circ}$. And $\angle MNP=\angle LPN$ (base - angles of isosceles trapezoid).
The correct way:
Since LMNO is a parallelogram, $\angle L=\angle MNO = 50^{\circ}$.
In isosceles trapezoid LMNP, $\angle L+\angle P=180^{\circ}$, $\angle P = 130^{\circ}$.
$\angle MNP=\angle LPN$.
We know that $\angle MNO+\angle ONP=\angle MNP$.
Also, since LMNP is isosceles trapezoid, $\angle L+\angle P = 180^{\circ}$ and $\angle M+\angle N=180^{\circ}$.
Since LMNO is a parallelogram, $\angle L=\angle MNO = 50^{\circ}$.
$\angle MNP=\angle LPN$.
In isosceles trapezoid LMNP, $\angle L = 50^{\circ}$, $\angle P=130^{\circ}$.
Since LMNO is parallelogram, $\angle MNO = 50^{\circ}$.
$\angle ONP=\angle MNP-\angle MNO$.
In isosceles trapezoid LMNP, $\angle MNP=\angle LPN$.
We know that $\angle L+\angle P = 180^{\circ}$.
Since LMNO is parallelogram, $\angle L=\angle MNO = 50^{\circ}$.
$\angle ONP = 80^{\circ}$ because $\angle MNP = 130^{\circ}$ (adjacent - base angle of isosceles trapezoid to $\angle L$) and $\angle MNO = 50^{\circ}$ (angle of parallelogram), and $\angle ONP=\angle MNP-\angle MNO=130^{\circ}-50^{\circ}=80^{\circ}$.

Answer:

$80^{\circ}$