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:

Step 1: Find the interior angle at Q

The exterior angle at Q and the interior angle at Q are supplementary (they add up to \(180^\circ\)). Given the exterior angle is \(155^\circ\), the interior angle \(\angle PQR = 180^\circ - 155^\circ = 25^\circ\).

Step 2: Use properties of a parallelogram

In a parallelogram, opposite angles are equal, and consecutive angles are supplementary. Also, if one pair of sides is equal and parallel, or if both pairs of opposite sides are equal, or if one pair of sides is equal and the corresponding angles are equal (implying parallelism), the quadrilateral is a parallelogram. Here, we know \(PS = QR\). For \(PQRS\) to be a parallelogram, \(\angle PSR\) should be equal to \(\angle PQR\) if we consider the sides \(PS\) and \(QR\) with the relevant angles, but wait, actually, in a parallelogram, consecutive angles are supplementary, but also, if \(PS = QR\) and we want \(PQRS\) to be a parallelogram, we need \(PS \parallel QR\). The angle \(\angle PSR\) and the interior angle at Q: Wait, the exterior angle at Q is \(155^\circ\), so the interior angle at Q is \(25^\circ\). For \(PS \parallel QR\), the alternate interior angles should be equal. Wait, maybe I made a mistake. Let's recall: If we extend \(PQ\) to get the exterior angle at Q, then the interior angle at Q is adjacent to that exterior angle. So \(\angle PQR = 180^\circ - 155^\circ = 25^\circ\). Now, in a parallelogram, \(PQ \parallel SR\) and \(PS \parallel QR\). So \(\angle PSR\) and \(\angle PQR\): Wait, no, in a parallelogram, opposite angles are equal. Wait, no, let's think about the sides. If \(PS = QR\) and we want \(PQRS\) to be a parallelogram, we need \(PS \parallel QR\). For \(PS \parallel QR\), the angle \(\angle PSR\) and the angle \(\angle PQR\) – wait, maybe the angle \(\angle PSR\) should be equal to the interior angle at Q? No, wait, actually, the consecutive angles in a parallelogram are supplementary, but also, if one pair of sides is equal and the corresponding angles (that would make the sides parallel) are equal. Wait, another approach: In a parallelogram, opposite angles are equal, and also, if a quadrilateral has one pair of sides equal and parallel, it's a parallelogram. Here, \(PS = QR\). If we can show that \(PS \parallel QR\), then with \(PS = QR\), \(PQRS\) is a parallelogram. For \(PS \parallel QR\), the alternate interior angles should be equal. The angle \(\angle PSR\) and the angle \(\angle PQR\) – wait, no, let's consider the transversal. Wait, maybe the angle \(\angle PSR\) should be equal to the interior angle at Q? Wait, no, the exterior angle is \(155^\circ\), so the interior angle at Q is \(25^\circ\). Wait, no, maybe I got the angle wrong. Wait, when you extend \(PQ\) beyond Q, the exterior angle at Q is formed. 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, in a parallelogram, \(PQ \parallel SR\) and \(PS \parallel QR\). So \(\angle PSR\) and \(\angle PQR\): Wait, no, \(\angle PSR\) and \(\angle PQR\) – if \(PS \parallel QR\) and \(PQ\) is a transversal, then \(\angle PSR\) and \(\angle PQR\) would be... Wait, maybe I mixed up the angles. Wait, actually, in a parallelogram, opposite angles are equal. So if \(PQRS\) is a parallelogram, then \(\angle PSR = \angle PQR\)? No, that can't be, because consecutive angles are supplementary. Wait, no, let's draw the quadrilateral: \(P - Q - R - S - P\). Extend \(PQ\) beyond Q to a point, say, T. So \(\angle RQT = 155^\circ\) (exterior angle at Q). Then \(\angle PQR = 180^\circ - 155^\circ = 25^\circ\). Now, for \(PQRS\) to be a parallelogram, \(PS \parallel QR\) and \(PQ \parallel SR\). If \(PS = QR\) and we want \(PS \parallel QR\), then the angle \(\angle PSR\) should be equal to \(\angle PQR\) if we consider the alternate interior angles? Wait, no, maybe the angle \(\angle PSR\) should be equal to \(155^\circ\)? Wait, no, that doesn't make sense. Wait, let's recall the property: In a parallelogram, consecutive angles are supplementary. So if \(\angle PQR = 25^\circ\), then the consecutive angle \(\angle QRS\) would be \(155^\circ\), and \(\angle PSR\) would be equal to \(\angle PQR\) if they are opposite? No, opposite angles are equal. So \(\angle P = \angle R\), \(\angle Q = \angle S\). Wait, maybe I labeled the quadrilateral wrong. Let's assume the quadrilateral is \(P, Q, R, S\) in order, so sides \(PQ\), \(QR\), \(RS\), \(SP\). Then extending \(PQ\) beyond Q (so from P to Q to T, where T is beyond Q), the exterior angle at Q is \(\angle RQT = 155^\circ\), so interior angle at Q is \(\angle PQR = 25^\circ\). Now, for \(PQRS\) to be a parallelogram, \(PS \parallel QR\) and \(PQ \parallel SR\). So \(PS \parallel QR\) implies that \(\angle PSR + \angle QRS = 180^\circ\) (consecutive angles), but also, \(PQ \parallel SR\) implies \(\angle PQR + \angle QRS = 180^\circ\) (consecutive angles). Therefore, \(\angle PSR = \angle PQR = 25^\circ\)? Wait, no, that would mean consecutive angles are equal, which would mean they are \(90^\circ\) if supplementary, but that's not the case. Wait, I think I made a mistake here. Let's start over.

The exterior angle at Q (when \(PQ\) is extended) is \(155^\circ\), so the interior angle at Q is \(180 - 155 = 25^\circ\). For \(PQRS\) to be a parallelogram, we need \(PS \parallel QR\) (since \(PS = QR\), if they are parallel, then it's a parallelogram). When \(PS \parallel QR\), the alternate interior angle to \(\angle PSR\) with respect to transversal \(SR\) would be... Wait, maybe the angle \(\angle PSR\) should be equal to the exterior angle? No, wait, the interior angle at Q is \(25^\circ\), so if \(PS \parallel QR\), then \(\angle PSR\) should be equal to \(180^\circ - 25^\circ = 155^\circ\)? Wait, no, let's use the property of parallelograms: In a parallelogram, opposite sides are equal and parallel, and opposite angles are equal, consecutive angles are supplementary. Also, if one pair of sides is equal and parallel, the quadrilateral is a parallelogram. Here, \(PS = QR\). If we can show \(PS \parallel QR\), then \(PQRS\) is a parallelogram. For \(PS \parallel QR\), the consecutive interior angles should be supplementary. Wait, the angle at S (\(\angle PSR\)) and the angle at Q (\(\angle PQR\)): If \(PS \parallel QR\) and \(PQ\) is a transversal, then \(\angle P + \angle Q = 180^\circ\), but maybe that's not the right transversal. Wait, the transversal for \(PS\) and \(QR\) would be \(SR\) or \(PQ\). Let's take transversal \(SR\): Then \(\angle PSR\) and \(\angle QRS\) are consecutive angles, so they should be supplementary. But also, with transversal \(PQ\), \(\angle PQR\) and \(\angle QRS\) are consecutive angles, so they should be supplementary. Therefore, \(\angle PSR = \angle PQR = 25^\circ\)? No, that would mean \(\angle PSR + \angle QRS = 180^\circ\) and \(\angle PQR + \angle QRS = 180^\circ\), so \(\angle PSR = \angle PQR\). But \(\angle PQR = 25^\circ\), so \(\angle PSR = 25^\circ\)? Wait, but that seems small. Wait, maybe I messed up the exterior angle. The exterior angle at Q when \(PQ\) is extended: so the side \(PQ\) is extended beyond Q, so the exterior angle is adjacent to \(\angle PQR\), so \(\angle PQR + \text{exterior angle} = 180^\circ\), so \(\angle PQR = 25^\circ\). Now, in a parallelogram, opposite angles are equal. So \(\angle PSR\) is opposite to \(\angle PQR\)? Wait, no, in quadrilateral \(PQRS\), the vertices are in order, so opposite angles are \(\angle P\) and \(\angle R\), \(\angle Q\) and \(\angle S\) (where \(\angle S\) is \(\angle PSR\)). So \(\angle Q = \angle S\), so \(\angle PSR = \angle PQR = 25^\circ\)? But that seems conflicting with the consecutive angles being supplementary. Wait, no, if \(\angle Q = 25^\circ\), then \(\angle S = 25^\circ\), and \(\angle P = \angle R = 155^\circ\), because consecutive angles are supplementary (\(25^\circ + 155^\circ = 180^\circ\)). Ah, that makes sense. So if \(\angle PSR = 25^\circ\), then since \(PS = QR\) and \(\angle PSR = \angle PQR\), and if we can show that \(PQ \parallel SR\) (because \(\angle P + \angle Q = 180^\circ\) if \(\angle P = 155^\circ\) and \(\angle Q = 25^\circ\)), then \(PQRS\) would be a parallelogram. Wait, but let's confirm: In a parallelogram, opposite angles are equal. So \(\angle Q = \angle S\), so \(\angle PSR = \angle PQR = 25^\circ\). Yes, that's the key. So the measure of \(\angle PSR\) that confirms \(PQRS\) is a parallelogram is \(25^\circ\).

Answer:

\(25^\circ\)