Question

triangle qrs is formed by connecting the midpoints of the side of triangle nop. the measures of the interior angles of triangle nop are shown. find the measure of ∠orq. figures not necessarily drawn to scale.

Explanation:

Step1: Find angle at NOP

In triangle \( NOP \), the sum of interior angles is \( 180^\circ \). Given \( \angle N = 75^\circ \), \( \angle P = 67^\circ \), so \( \angle O = 180 - 75 - 67 = 38^\circ \). Wait, no, the angle at \( O \) is given as \( 38^\circ \)? Wait, the diagram shows angle at \( O \) is \( 38^\circ \), angle at \( N \) is \( 75^\circ \), angle at \( P \) is \( 67^\circ \). Wait, actually, first, since \( Q, R, S \) are midpoints, by Midline Theorem, \( QR \parallel NP \), \( QS \parallel OP \), \( SR \parallel NO \). So triangle \( QRS \) is similar to triangle \( NOP \), and also, \( \triangle OQR \) has \( OQ \) as part of \( NO \), \( OR \) as part of \( OP \)? Wait, no, let's re-examine.

Wait, the problem is to find \( \angle ORQ \). Let's first find the third angle of \( \triangle NOP \). Sum of angles in a triangle: \( \angle N + \angle O + \angle P = 180^\circ \). So \( 75^\circ + \angle O + 67^\circ = 180^\circ \). So \( \angle O = 180 - 75 - 67 = 38^\circ \). Now, since \( Q \) and \( R \) are midpoints, \( QR \) is the midline of \( \triangle NOP \), so \( QR \parallel NP \). Therefore, \( \angle ORQ = \angle P \) (corresponding angles, because \( QR \parallel NP \) and \( OP \) is a transversal). Wait, \( \angle P = 67^\circ \)? Wait, no, let's check again.

Wait, midline theorem: the segment connecting midpoints of two sides is parallel to the third side. So \( Q \) is midpoint of \( NO \), \( R \) is midpoint of \( OP \), so \( QR \parallel NP \). Therefore, \( \angle ORQ \) and \( \angle P \) are corresponding angles, so \( \angle ORQ = \angle P = 67^\circ \)? Wait, no, maybe I mixed up. Wait, \( Q \) is midpoint of \( NO \), \( S \) is midpoint of \( NP \), \( R \) is midpoint of \( OP \). So \( QR \) connects midpoints of \( NO \) and \( OP \), so \( QR \parallel NP \). Therefore, \( \angle ORQ \) and \( \angle P \) are corresponding angles, so \( \angle ORQ = \angle P = 67^\circ \)? Wait, but let's check the angle at \( O \). Wait, \( \angle O = 38^\circ \), as calculated. Wait, maybe another approach: since \( R \) is midpoint of \( OP \), and \( Q \) is midpoint of \( NO \), then \( QR \parallel NP \), so \( \angle ORQ = \angle P \), because \( QR \parallel NP \), so corresponding angles. \( \angle P = 67^\circ \), so \( \angle ORQ = 67^\circ \)? Wait, but let's verify the sum of angles in \( \triangle NOP \): \( 75 + 38 + 67 = 180 \), yes, \( 75 + 38 = 113 \), \( 113 + 67 = 180 \). So that's correct. Now, since \( QR \parallel NP \), \( \angle ORQ = \angle P = 67^\circ \). Wait, but maybe I made a mistake. Alternatively, \( \triangle OQR \): \( OQ = \frac{1}{2}NO \), \( OR = \frac{1}{2}OP \), so \( \triangle OQR \sim \triangle ONP \) by SAS similarity (ratio 1/2, included angle \( \angle O \) common). Therefore, \( \angle ORQ = \angle P = 67^\circ \).

Step2: Confirm using midline theorem

Midline theorem: segment joining midpoints of two sides is parallel to the third side and half its length. So \( QR \parallel NP \), so \( \angle ORQ \) and \( \angle P \) are corresponding angles, hence equal. Therefore, \( \angle ORQ = 67^\circ \).

Answer:

\( 67^\circ \)