Question

can you conclude that \\(\overline{hp}\\) and \\(\overline{kn}\\) are congruent?

Explanation:

Step1: Identify the polygon type

The figure is a hexagon. Looking at the markings, sides \( KJ \), \( IH \), \( HP \), and \( KN \)? Wait, no, let's check the congruency marks. Sides with two marks: \( KJ \), \( IH \), \( HP \), \( KN \)? Wait, actually, in a regular hexagon, all sides are congruent, but here the markings: \( JI \) has one mark, \( NP \) has one mark, \( KJ \) has two marks, \( IH \) has two marks, \( HP \) has two marks, \( KN \) has two marks? Wait, no, the diagram: \( JI \) (top) has one tick, \( NP \) (bottom) has one tick. \( KJ \) (left top) has two ticks, \( IH \) (right top) has two ticks, \( HP \) (right bottom) has two ticks, \( KN \) (left bottom) has two ticks. So sides with two ticks: \( KJ \cong IH \cong HP \cong KN \)? Wait, no, let's count the sides. A hexagon has 6 sides: \( JI \), \( IH \), \( HP \), \( PN \), \( NK \), \( KJ \). Wait, maybe I mislabeled. Wait the vertices are \( J, I, H, P, N, K \) (connecting \( J \) to \( I \), \( I \) to \( H \), \( H \) to \( P \), \( P \) to \( N \), \( N \) to \( K \), \( K \) to \( J \)). So sides: \( JI \) (1 tick), \( IH \) (2 ticks), \( HP \) (2 ticks), \( PN \) (1 tick), \( NK \) (2 ticks), \( KJ \) (2 ticks). So sides with two ticks: \( IH \), \( HP \), \( NK \), \( KJ \). So \( IH \cong HP \cong NK \cong KJ \). Wait, \( NK \) is \( KN \) (since \( N \) to \( K \) is same as \( K \) to \( N \)). So \( HP \) and \( KN \) (which is \( NK \)) are both sides with two ticks, so they are congruent. Wait, but maybe the hexagon is equilateral (all sides with same tick marks are congruent). So sides with two ticks are congruent to each other. So \( HP \) (two ticks) and \( KN \) (two ticks) should be congruent.

Step2: Conclude congruency

Since \( \overline{HP} \) and \( \overline{KN} \) are both marked with two congruency ticks (indicating they are part of the set of congruent sides), we can conclude they are congruent.

Answer:

yes