Question

in quadrilateral pqrs, the exterior angle at q when \\(\overline{pq}\\) is extended measures 155 and \\(ps = qr\\). what measure of \\(\angle psr\\) would confirm that pqrs is a parallelogram? \\(m\angle psr = \square\\)

Explanation:

Step1: Find the interior angle at Q

An exterior angle and its adjacent interior angle are supplementary (sum to \(180^\circ\)). The exterior angle at Q is \(155^\circ\), so the interior angle \(\angle PQR = 180^\circ - 155^\circ = 25^\circ\)? Wait, no, wait. Wait, if we extend \(\overline{PQ}\) to form the exterior angle at Q, the exterior angle and the interior angle at Q (which is \(\angle PQR\)) are supplementary. Wait, actually, when you extend a side of a polygon, the exterior angle and the adjacent interior angle are supplementary. So if the exterior angle is \(155^\circ\), then the interior angle at Q (let's say \(\angle PQR\)) is \(180^\circ - 155^\circ = 25^\circ\)? Wait, no, that can't be right. Wait, maybe I got the direction wrong. Wait, in a quadrilateral, when you extend a side, the exterior angle is supplementary to the interior angle at that vertex. So if we extend \(\overline{PQ}\) beyond Q to form a ray, then the exterior angle is between the extended \(\overline{PQ}\) and \(\overline{QR}\). So the interior angle at Q ( \(\angle PQR\)) and the exterior angle are supplementary. So \(\angle PQR + 155^\circ = 180^\circ\), so \(\angle PQR = 25^\circ\)? Wait, but for a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Wait, but we also know that \(PS = QR\). In a parallelogram, opposite sides are equal, so if \(PS = QR\), and if we can show that \(PS \parallel QR\) and \(PQ \parallel SR\), then it's a parallelogram. Alternatively, if one pair of sides is equal and parallel, then it's a parallelogram. Wait, but maybe we need to find the measure of \(\angle PSR\) such that \(PQ \parallel SR\), so that consecutive angles are supplementary. Wait, let's re-examine.

Wait, the exterior angle at Q is \(155^\circ\), so the interior angle at Q ( \(\angle PQR\)) is \(180^\circ - 155^\circ = 25^\circ\)? No, that seems small. Wait, maybe I mixed up the exterior angle. Wait, no, when you extend a side, the exterior angle is equal to the sum of the two non-adjacent interior angles (by the exterior angle theorem), but in a quadrilateral, maybe that's not the case. Wait, no, in any polygon, the exterior angle and the adjacent interior angle are supplementary. So if the exterior angle is \(155^\circ\), the adjacent interior angle (at Q) is \(180 - 155 = 25^\circ\). But in a parallelogram, consecutive angles are supplementary. Wait, but if PQRS is a parallelogram, then \(\angle PSR\) should be equal to \(\angle PQR\) if they are opposite angles, but that would be \(25^\circ\), which seems odd. Wait, maybe I made a mistake here. Wait, maybe the exterior angle is actually the angle we get when we extend \(\overline{QP}\) instead of \(\overline{PQ}\). Let's clarify: the problem says "the exterior angle at Q when \(\overline{PQ}\) is extended". So we extend \(\overline{PQ}\) beyond Q, so the side is PQ, extended to, say, point T, so QT is the extension. Then the exterior angle is \(\angle RQT = 155^\circ\). Then the interior angle at Q is \(\angle PQR\), which is adjacent to \(\angle RQT\), so they are supplementary. So \(\angle PQR + \angle RQT = 180^\circ\), so \(\angle PQR = 180 - 155 = 25^\circ\). But in a parallelogram, opposite angles are equal, so \(\angle PSR = \angle PQR\) if they are opposite angles. Wait, but let's check the sides: \(PS = QR\). In a parallelogram, opposite sides are equal, so \(PS = QR\) and \(PQ = SR\). Also, consecutive angles are supplementary. Wait, maybe I got the exterior angle wrong. Wait, maybe the exterior angle is \(155^\circ\), so the interior angle is \(180 - 155 = 25^\circ\), but that seems too small. Wait, maybe the exterior angle is actually \(155^\circ\), so the interior angle is \(180 - 155 = 25^\circ\), but in a parallelogram, consecutive angles are supplementary. Wait, no, consecutive angles in a parallelogram are supplementary. So if \(\angle PQR\) is \(25^\circ\), then the consecutive angle \(\angle QRS\) would be \(155^\circ\), and \(\angle PSR\) would be equal to \(\angle PQR\) (opposite angles). Wait, this is confusing. Wait, maybe I made a mistake in the exterior angle. Wait, let's start over.

In a quadrilateral, to be a parallelogram, opposite sides must be parallel and equal, or one pair of sides is parallel and equal, or opposite angles are equal, or consecutive angles are supplementary. We know that \(PS = QR\). If we can show that \(PS \parallel QR\), then since \(PS = QR\), the quadrilateral would be a parallelogram (one pair of sides parallel and equal). For \(PS \parallel QR\), the consecutive interior angles (at S and Q) should be supplementary? Wait, no, if \(PS \parallel QR\), then the alternate interior angles would be equal, or consecutive interior angles would be supplementary. Wait, the transversal would be \(SR\) or \(PQ\). Wait, maybe the exterior angle at Q is \(155^\circ\), so the angle between \(PQ\) (extended) and \(QR\) is \(155^\circ\), so the angle between \(PQ\) (original) and \(QR\) is \(25^\circ\) (since they are supplementary). Then, if \(PQ \parallel SR\), then the angle at S ( \(\angle PSR\)) and the angle at Q ( \(\angle PQR\)) would be same-side interior angles, hence supplementary. Wait, that makes more sense. So if \(PQ \parallel SR\), then \(\angle PSR + \angle PQR = 180^\circ\). Wait, but we found \(\angle PQR = 25^\circ\), so \(\angle PSR = 155^\circ\)? Wait, that contradicts the earlier thought. Wait, let's clarify the exterior angle.

Wait, the exterior angle is formed by extending \(PQ\) beyond Q, so the exterior angle is adjacent to \(\angle PQR\). So \(\angle PQR + \text{exterior angle} = 180^\circ\), so \(\angle PQR = 180 - 155 = 25^\circ\). Now, if PQRS is a parallelogram, then \(PQ \parallel SR\), so \(\angle PSR\) and \(\angle PQR\) are same-side interior angles, hence supplementary. Therefore, \(\angle PSR + \angle PQR = 180^\circ\), so \(\angle PSR = 180 - 25 = 155^\circ\). Wait, that makes sense. Because in a parallelogram, consecutive angles are supplementary. So if \(PQ \parallel SR\), then \(\angle PSR + \angle PQR = 180^\circ\). Since \(\angle PQR = 25^\circ\), then \(\angle PSR = 155^\circ\). Alternatively, maybe the exterior angle is equal to the opposite interior angle? No, that's in a triangle. Wait, in a parallelogram, opposite angles are equal, and consecutive angles are supplementary. So if the exterior angle at Q is \(155^\circ\), then the interior angle at Q is \(25^\circ\), so the consecutive angle at S ( \(\angle PSR\)) should be \(180 - 25 = 155^\circ\) to make them supplementary, hence \(PQ \parallel SR\). And since \(PS = QR\), if \(PS \parallel QR\) (which would be the case if \(PQ \parallel SR\) and it's a parallelogram), then it's a parallelogram. Wait, but we also have \(PS = QR\). So if \(PS \parallel QR\) and \(PS = QR\), then PQRS is a parallelogram. So to confirm it's a parallelogram, we need \(PS \parallel QR\), which requires that \(\angle PSR + \angle PQR = 180^\circ\) (consecutive interior angles). Since \(\angle PQR = 25^\circ\), then \(\angle PSR = 155^\circ\). That must be it. So the measure of \(\angle PSR\) is \(155^\circ\).

Step1: Calculate the interior angle at Q

The exterior angle at Q is \(155^\circ\), and an exterior angle and its adjacent interior angle are supplementary. So:
\[ \angle PQR = 180^\circ - 155^\circ = 25^\circ \]

Step2: Determine \(\angle PSR\) for parallelogram

In a parallelogram, consecutive angles are supplementary (since opposite sides are parallel). If \(PQ \parallel SR\), then \(\angle PSR\) and \(\angle PQR\) are consecutive angles (same - side interior angles) and thus supplementary:
\[ \angle PSR + \angle PQR = 180^\circ \]
Substitute \(\angle PQR = 25^\circ\) into the equation:
\[ \angle PSR = 180^\circ - 25^\circ = 155^\circ \]

Answer:

\(155^\circ\)