Note that the first step of this construction does not use any of the hypotheses. After these three reflections, the triangle $ABC$ has been moved on top of triangle $DEF$ so the two are congruent. Since segment $DF$ is congruent to segment $AC$ by hypothesis and $AC$ is congruent to $DC''$ (because reflections preserve lengths of line segments) the reflection about line $DE$ maps $C''$ to $F$. Since all rigid motions of the plane preserve angles, angle $C''DE$ must map to angle $FDE$. Angle $FDE$ is congruent to angle $C''DE$ because angle $C''DE$ is the image under two reflections of angle $CAB$ which is congruent to angle $FDE$ by hypothesis. So we have to check that reflection about line $DE$ maps $C''$ to $F$. The only reflection that leaves $D$ and $E$ fixed is the one about line $DE$. In this last step we must move $C''$ to $F$ via a reflection while leaving $D$ and $E$ fixed. The result of the second reflection is pictured below: The reason we know that $D$ is on the perpendicular bisector of $B'E$ is that it is equidistant from $E$ and $B'$ by the hypothesis that $AB$ is congruent to $DE$ and the perpendicular bisector of a line segment $xy$ consists of all points in the plane equidistant from $x$ and $y$. Note that it is important that this perpendicular bisector contains $D$ so that our second reflection preserves what we accomplished in the first step. In this step we wish to move $B'$ to $E$ and so we must reflect again, this time about the perpendicular bisector of $B'E$. Also pictured below is the new triangle $DB^\prime C^\prime$ obtained by reflecting triangle $ABC$. So we must reflect about the perpendicular bisector of $AD$ which is pictured below. In the first part of this problem, we wish to send $A$ to $D$ via a reflection. Reflection about line $L$ sends point $P$ in the plane to point $Q$ exactly when $L$ is the perpendicular bisector of $PQ$
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |