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{\Large \bf MATH 476 -- College Geometry}
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\noindent
{\large \bf Solutions to Homework Assignment 2}

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\noindent {\bf Section 2.6:} 1, 4, 5, 10
\be
\item[1.]
\be \item $H(P,\lin{TS})$
    \item $H(T,\lin{PR})$
    \item $H(P,\lin{TS}) \cap H(T,\lin{PR})$
\ee
\item[4.] Since $\lin{BC}$ partitions the plane, the ``pieces'' are disjoint or equal.  If $D\in H(A,\lin{BC}),$ then
$H(D,\lin{BC})$ and $H(A,\lin{BC})$ are clearly not disjoint; thus, they must be equal.
\item[5.] \bob
\item[10.] This is false: suppose that the two lines are parallel.  Then the intersection of the two half-planes that meet is a
strip, not an angle interior.
\ee

\noindent {\bf Section 3.1:} 2, 4, 8, 11, 12, 13
\be
 \item[2.]
 \be
  \item[(a)] $\angle ACB\cong\angle ACD.$  $\angle SVU\cong\angle RVW.$
  \item[(b)] $m\angle UVR=m\angle SVU+m\angle SVR= m\angle RVW+m\angle SVR =m\angle SVW.$
 \ee

 \item[4.] $\ov{LN}\cong \ov{RT}, \ov{rs}\cong\ov{LM}, \ov{ST}\cong\ov{MN}.$

 \item[8.] $\triangle ABC$ and $\triangle FED$ satisfy the SAS hypothesis.
 \item[11.] \bob
 \item[12.] $\angle BAD=\ray{AD}\cup\ray{AB}.$  $\angle CAD=\ray{AD}\cup\ray{AC}.$ By Theorem 4 in Section 2.4,
      $\ray{AC}=\ray{AB},$ so these two sets are equal.  That is, $\angle BAD =\angle CAD,$ so
      they are congruent.
 \item[13.] \bob
\ee

\noindent {\bf Section 3.2:} 12
\be
\item[12.] Consider the points at $(0,0),(2,2),$ and $(2,-2).$  These points lie on infinitely many circles: any circle of radius $a$
centered at $(a,0)$ with $a\ge 1$ will pass through all three points.
\ee


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