Question

5 right triangle mnp is shown below. side $overline{np}$ lies on line $l$.
(a) using only a straightedge and a compass, find the image of point m after a reflection in line $l$. label it as $m$.
(b) give an explanation for why point m, n and $m$ must be collinear.
(c) $\triangle pmm$ must be isosceles (see problem 2(c)). angles $angle m$ and $angle m$ are called the base angles of the triangle. what must be true about these angles? use facts about rigid motions to justify your answer.

Explanation:

Part (b)

Step1: Recall Reflection Properties

A reflection over a line \( l \) maps a point \( M \) to \( M' \) such that \( l \) is the perpendicular bisector of \( \overline{MM'} \). Also, for a point \( N \) on line \( l \), the reflection of \( N \) over \( l \) is itself (\( N' = N \)).

Step2: Analyze Collinearity

Since \( N \) is on line \( l \), \( N' = N \). For three points \( M \), \( N \), \( M' \), we know that line \( l \) is the perpendicular bisector of \( \overline{MM'} \), so \( N \) (on \( l \)) lies on the line that is the perpendicular bisector of \( \overline{MM'} \). By the definition of reflection, the line \( l \) contains the midpoint of \( \overline{MM'} \) and is perpendicular to it. So, points \( M \), \( N \), and \( M' \) must be collinear because \( N \) is on the line of reflection (which is related to the segment \( \overline{MM'} \) by the reflection property), meaning they lie on a straight line.

Step1: Recall Isosceles Triangle and Rigid Motions

An isosceles triangle has two congruent sides, and the base angles (angles opposite the congruent sides) are congruent. A reflection is a rigid motion, which preserves distances and angles.

Step2: Analyze Angles \( \angle M \) and \( \angle M' \)

In \( \triangle PNM' \) (or considering the reflection of \( \triangle PNM \) over line \( l \)), since \( M' \) is the reflection of \( M \) over \( l \), \( \overline{PM} = \overline{PM'} \) (reflection preserves distance) and \( \overline{NM} = \overline{NM'} \) (same reason). Also, \( \overline{PN} \) is common (or congruent by reflection). By the Side - Side - Side (SSS) congruence criterion (from rigid motion preservation of distance), \( \triangle PNM \cong \triangle PNM' \). Then, by the Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem, \( \angle M \cong \angle M' \), meaning \( \angle M \) and \( \angle M' \) are equal in measure.

Answer:

Points \( M \), \( N \), and \( M' \) are collinear because \( N \) lies on the line of reflection \( l \), and a reflection over \( l \) implies \( l \) is the perpendicular bisector of \( \overline{MM'} \), so \( N \) (on \( l \)) is on the line containing \( M \) and \( M' \) (due to the reflection's effect on the segment \( \overline{MM'} \)).

Part (c)