Solutions to in-class problems
Auteur:
Kathryn Bragwell
Last Updated:
il y a 9 ans
License:
Creative Commons CC BY 4.0
Résumé:
Solutions to in-class problems
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
Solutions to in-class problems
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[12pt]{article}
% * <keg2013@gmail.com> 2016-01-23T04:06:18.156Z:
%
% ^.
\usepackage[margin=1in]{geometry}
\usepackage{amsmath,amsthm,amssymb,amsfonts}
\usepackage{setspace}
\usepackage{tikz}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newenvironment{problem}[2][Problem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
\newenvironment{theorem}[2][Theorem]{\begin{trivlist}
\item[\hskip \labelsep {\bfseries #1}\hskip \labelsep {\bfseries #2.}]}{\end{trivlist}}
%If you want to title your bold things something different just make another thing exactly like this but replace "problem" with the name of the thing you want, like theorem or lemma or whatever
\begin{document}
\setstretch{3}
%\renewcommand{\qedsymbol}{\filledbox}
%Good resources for looking up how to do stuff:
%Binary operators: http://www.access2science.com/latex/Binary.html
%General help: http://en.wikibooks.org/wiki/LaTeX/Mathematics
%Or just google stuff
\title{Solutions to in-class problems}
\author{College Geometry}
\date{Spring 2016}
\maketitle
\begin{theorem}{3.1.7}
If $\ell$ and $m$ are two distinct, nonparallel lines, then there exists exactly one point $P$ such that $P$ lies on both $\ell$ and $m$.
\end{theorem}
\begin{proof}
(Selena Emerson) Let $\ell$ and $m$ be two distinct, nonparallel lines. By definition of nonparallel there exists at least one point, $P$, that lies on both $\ell$ and $m$. Suppose there exists a different point, $Q$, be on both $\ell$ and $m$. By Axiom 3.1.3, there exists exactly one line on which the two distinct points lie. Since $P$ and $Q$ are on both $\ell$ and $m$, then $\ell$ and $m$ are the same line. But $\ell$ and $m$ are defined as being distinct lines which is a contradiction. Thus, there exists exactly one point $P$ such that $P$ lies on both $\ell$ and $m$.
\end{proof}
\begin{problem}{3.2.3}
Show that the taxicab metric defined in Example 3.2.11 is a metric(i.e., verify that the function $\rho$ satisfies the three conditions in the definition of metric on page 339).
\end{problem}
\begin{proof}
\textit{(Brianna Hillman)} We want to show that the taxicab metric is a metric. To show this, we need to show $D(P,Q) = D(Q,P)$ for every $P$ and $Q$, $D(P,Q) \geq 0$ for every $P$ and $Q$, and that $D(P,Q) = 0$ if and only if $P$ = $Q$. Let $P = (x_{1},y_{1})$ and $Q = (x_{2},y_{2})$. Take $P, Q \in R^2$. Then we have that
$\rho ((x_{1},y_{1}), (x_{2},y_{2})) = \rho ((x_{2},y_{2}), (x_{1},y_{1}))$ \\
$\Rightarrow \mid x_{2} - x_{1} \mid + \mid y_{2}- y_{1} \mid = \mid x_{1} - x_{2} \mid + \mid y_{1}- y_{2} \mid$. \\ We know that this is equal because of the properties of the absolute values. Thus, the first condition of a metric is satisfied. Now, take a $P$, $Q$ $\in R^2$. If $P$ = $Q$, then \\ $\rho ((x_{1},y_{1}), (x_{1},y_{1})) = \mid x_{1} - x_{1} \mid + \mid y_{1}- y_{1} \mid = 0$. But, if $P \neq Q$, then \\ $PQ = \mid x_{2} - x_{1} \mid + \mid y_{2} - y_{1} \mid \geq 0$ . Thus, $\rho ((x_{1},y_{1}), (x_{2}, y_{2})) = \mid x_{2} - x_{1} \mid + \mid y_{2}- y_{1} \mid \geq 0$. Therefore, the second condition of a metric is satisfied. Next we will prove the third condition. ($\Rightarrow$) Suppose $\rho (P,Q) = 0$. Then, $\mid x_{2} - x_{1} \mid + \mid y_{2}- y_{1} \mid = 0$. So $\mid x_{2} - x_{1} \mid = 0$ and $\mid y_{2} - y_{1} \mid = 0$. Therefore, $x_{2} = x_{1}$ and $y_{2} = y_{1}$ or $P$ = $Q$. ($\Leftarrow$) Now, suppose $P$ = $Q$. So, $P = Q = (x_{1},y_{1})$. Then, $D (P,Q) = \rho ((x_{1},y_{1}), (x_{1},y_{1}))$ \\
$\Rightarrow = \mid x_{1} - x_{1} \mid + \mid y_{1}- y_{1} \mid$ \\
$\Rightarrow = 0 + 0$ \\
$\Rightarrow = 0$. \\
Therefore, the third condition of a metric is satisfied and the taxicab metric is a metric.
\end{proof}
\begin{problem}{3.2.7}
Find all point $(x,y)$ in $\mathbb{R}^2$ such that $\rho((0,0),(x,y)) = 1$, where $\rho$ is the taxi cab metric. Draw a sketch in the Cartesian plane. (This shape might be called a "circle" in the taxicab metric.)
\end{problem}
\begin{proof} \textit{(Victoria Krohn)}
$\rho((0,0),(x,y)) = |x-0|+|y-0|= 1$\\
$(x,y) = (1,0)$ \hspace{1cm} $\rho((0,0),(1,0)) = |1-0|+|0-0|= 1$\\
$(x,y) = (0,1)$ \hspace{1cm} $\rho((0,0),(0,1)) = |0-0|+|1-0|= 1$\\
$(x,y) = (-1,0)$ \hspace{1cm} $\rho((0,0),(-1,0)) = |(-1)-0|+|0-0|= 1$\\
$(x,y) = (0,-1)$ \hspace{1cm} $\rho((0,0),(0,-1)) = |0-0|+|(-1)-0|= 1$\\
\begin{tikzpicture}
\draw[step=1cm,gray,very thin] (-2,-2) grid (2,2);
\foreach \x in {-2,-1,0,1,2}
\draw (\x cm,1pt) -- (\x cm,-1pt) node[anchor=north] {$\x$};
\foreach \y in {-2,-1,0,1,2}
\draw (0pt,\y cm) -- (0 pt,\y cm) node[anchor=east] {$\y$};
\draw (-1,0) -- (0,1) -- (1,0) -- (0,-1) -- (-1,0);
\end{tikzpicture}
\end{proof}
\begin{problem}{3.2.21}
Let \textit{A} and \textit{B} be two distinct points. Prove that $\overline{AB}$ = $\overline{BA}$.
\end{problem}
\begin{proof}
\textit{(Brianna Hillman)} Assume A and B are distinct points. Note that \\ $\overline{AB} = \{A,B\} \cup \{P \mid A \star P \star B\}$ and $\overline{BA} = \{B,A\} \cup \{P \mid B \star P \star A\}$. Let $x \in \overline{AB}$. \\ So $x \in \{A,B\} \cup \{P \mid A \star P \star B\}$ \\
$\Rightarrow x \in \{A,B\}$ or $x \in \{P \mid A \star P \star B\}$ \\
In the case where $x = A$ or $x = B$, then $x \in \overline{BA}$. In the case where $x \neq A$ and $x \neq B$, $x$ is in between A and B. Therefore, $Ax + xB = AB$ by the definition of between. Since $AB = Ax + xB$, then we have that \\
$\Rightarrow AB = xB + Ax$ \\
$\Rightarrow AB = Bx + xA$ by Theorem 3.2.7 \\
Since AB = Bx + xA, then x is between B and A and hence, $\overline{AB} \subseteq \overline{BA}$. Similarly, if we take an $x \in \overline{BA}$, we conclude that $BA = Ax + xB$ and $x$ is between A and B. Thus, $\overline{BA} \subseteq \overline{AB}$. Therefore, $\overline{AB} = \overline{BA}$.
\end{proof}
\begin{theorem}{3.3.12}(Pasch's Axiom)
Let $\triangle {ABC}$ be a triangle and let \textit{l} be a line such that none of A, B, and C lies on \textit{l}. If \textit{l} intersects $\overline{AB}$, then \textit{l} also intersects BC or AC.
\end{theorem}
\begin{proof}
(Katelynn Gordon) Let points A, B, and C form a triangle $\triangle$ABC and let \textit{l} be a line that does not go through a vertex of $\triangle$ABC. Then \textit{l} has to go in and come out of the triangle. Then there are three options: \textit{C} $\in$ \textit{l}, \textit{C} $\in$ \textit{$H_1$}, or \textit{C} $\in$ \textit{$H_2$}. If \text{C} $\in$ \textit{$H_1$}, then since \textit{B} $\in$ \textit{$H_2$}, $\overline{BC}$ $\cap$ \textit{l} $\neq$ $\emptyset$ (3.3.4). Or if \textit{C} $\in$ \textit{$H_2$}, then since \textit{A} $\in$ \textit{$H_1$} $\overline{AC}$ $\cap$ \textit{l} $\neq$ $\emptyset$. In either case, \textit{l} crosses $\overline{AC}$ or $\overline{BC}$.
\end{proof}
\begin{problem}{3.3.3}
Let $l$ be a line and let $H$ be one of the half-planes bounded by $l$. Prove that $H \cup l$ is a convex set.
\end{problem}
\begin{proof}
\textit{(Victoria Krohn)} Let $l$ be a line and $H$ be a half-plane bounded by $l$. We need to show that that $H \cup l$. Let points $A,B$ exist such that they construct $\overline{AB}$. We consider 3 cases, when $\overline{AB} \cup H$, when $\overline{AB} \cup l$ and when $A \cup l$ and $B \cup H$. When $\overline{AB} \cup H$ by the Plane Separation Postulate $\overline{AB}$ is in $H \cup l$. When $\overline{AB} \cup l$ by the Incidence Postulate they are in $H \cup l$. Let $A \in l$ and $B \in H$ so $\overrightarrow{AB} \in H$ by the Ray Theorem, therefore $H \cup l$. Thus $H \cup l$ is convex.
\end{proof}
\begin{problem}{3.3.5}
Suppose $\triangle {ABC}$ is a triangle and $l$ is a line such that none of the vertices $A, B,$ or $C$ lies on $l$. Prove that $l$ cannot intersect all three sides of $\triangle {ABC}$. Is it possible for a line to intersect all three sides of a triangle?
\end{problem}
\begin{proof}
\textit{(Victoria Krohn)} Assume line $l$ intersects a side of a triangle $\triangle {ABC}$. Without loss of generality, $l$ intersects with side $\overline{AB}$. This creates 2 half-planes, $H_1, H_2$ with $A \in H_1$ and $B \in H_2$. By Pasch's Axiom, \textit{l} must exit $\triangle {ABC}$ through another side. Thus $C \in H_1$ or $C \in H_2$. Therefore, $l$ cannot cross all 3 sides of the triangle.
\end{proof}
\begin{problem}{3.5.1}
Prove: If $l \perp m$, then $l$ and $m$ contain rays that make four different right angles.
\end{problem}
\begin{proof}
\textit{Victoria Krohn} Let $A, B, C, D, E$ be distinct points, such that $A,B,E \in \text{line } l$ and $D,A,C \in \text{line }m$ with $m \perp l$ intersecting at point $A$. We need to show that $\mu\angle (BAC),$ $\mu\angle (CAE),$ $\mu\angle (DAE),$ and $\mu\angle (DAB)$ equal $90^\circ$.
By definition of perpendicular, $\angle BAC$ is a right angle, thus $\mu\angle (BAC) = 90^\circ$. Angles $\angle DAB$ and $\angle BAC$ are a linear pair by the opposite rays $\overrightarrow{AD}$ and $\overrightarrow{AC}$ (definition of linear pair). By the Linear Pair Theorem, $\mu\angle (BAC) + \mu\angle (DAB) = 180^\circ$. So, $90^\circ + \mu\angle (DAB) = 180^\circ$, $\mu\angle (DAB) = 90^\circ$. By definition of right angle, $\angle DAB$ is a right angle.
Similarly for angles $\angle DAB$ and $\angle DAE$, and $\angle DAE$ and $\angle CAE.$
\end{proof}
\begin{problem}{3.5.2}
Prove existence and uniqueness of a perpendicular to a line at a point on the line (Theorem 3.5.9).
\end{problem}
\begin{proof}{(Brianna Hillman)}
Let $\ell$ be a line where $P$ and $Q$ are two distinct points on $\ell$. By the Angle Construction Postulate, there exists a unique ray $\overrightarrow{PA}$ such that $A$ is in one half-plane bounded by $\ell$ and $\mu (\angle APQ) = 90^\circ$. Then, we can extend ray $\overrightarrow{PA}$ to a line by the Incidence Postulate. So, there exists exactly one line $\overleftrightarrow{PA}$ such that $P$ lies on $\overleftrightarrow{PA}$ and $\overleftrightarrow{PA} \perp \ell$.
\end{proof}
\begin{problem}{3.5.5}
Restate the Vertical Angles Theorem (Theorem 3.5.13) in if-then form. Prove the theorem.
\end{problem}
If the angles are vertical, then they are congruent.
\begin{proof}{(Selena Emerson)} Let $\angle{BAC}$ and $\angle{DAE}$ be vertical angles with $A$ being the intersection of $\overleftrightarrow{BE}$ and $\overleftrightarrow{CD}$. Then $\angle{BAC}$ and $\angle{CAE}$ are linear pairs, making $\mu\angle (BAC) + \mu\angle (CAE) = 180^\circ$ by the linear pair theorem. The same is true for $\angle{CAE}$ and $\angle{DAE}$. Thus, $\mu\angle (BAC) + \mu\angle (CAE) = \mu\angle (CAE) + \mu\angle (DAE)$ so $\mu\angle (BAC) = \mu\angle (DAE)$. Hence, $\angle{BAC} \cong \angle{DAE}$.
\end{proof}
\begin{problem}{3.5.6}
Prove the following coverse of the Vertical Angles Theorem: If A, B, C, D, and E, are points such that A*B*C, D and E are on opposite sides of $\leftrightarrow{AB}$, and $\angle{DBC} \cong \angle{ABE}$, then D, B, and E are collinear.
\end{problem}
\begin{proof}{Katelynn Gordon}
$BWOC$ Let If A, B, C, D, and E, are points such that A*B*C, D and E are on opposite sides of $\leftrightarrow{AB}$, and $\angle{DBC} \cong \angle{ABE}$. Then without loss of generality let $\overrightarrow{EB}$ be between $\angle{ABF}$. Then $\angle{DBC} \cong \angle{ABF}$ by the Veritcal Angles Theorem. Then $\mu (\angle ABE)$ + $\mu (\angle EBF)$ should equal $\mu (\angle DBC)$ by angle addition. However, we assumed $\angle{ABE} \cong \angle{DBC}$ from the problem. So $\mu (\angle EBF)$ would have to be 0$^{\circ}$ so $E$, $B$, and $D$ would have to be on the same line.
\end{proof}
\begin{theorem}{3.6.5 (Isosceles Triangle Theorem)}
The base angles of an isosceles triangle are congruent.
\end{theorem}
\begin{proof}{(Brianna Hillman)}
Let $\triangle ABC$ be a triangle such that $\overline{AB} \cong \overline{AC}$. We must prove that $\angle ABC \cong \angle ACB$. Let $D$ be a point in the interior of $\angle BAC$ such that $\overrightarrow{AD}$ is the bisector of $\angle BAC$ (Theorem 3.4.7). There is a point $E$ at which the ray $\overrightarrow{AD}$ intersects the segment $\overline{BC}$ (Crossbar Theorem). Then $\triangle BAE \cong \triangle CAE$ by SAS and so $\angle ABE \cong \angle ACE$. Thus $\angle ABC \cong \angle ACB$.
\end{proof}
\begin{problem}{3.7.1}
Check that the trivial geometry containing just one point and no lines satisfies all the postulates for neutral geometry except the Existence Postulate. Which parallel postulate is satisfied by this geometry?
\end{problem}
\begin{proof}
(Brianna Hillman) The Ruler Postulate and Incidence Postulate are vacuously true because there is only one point in trivial geometry. The Plane Separation Postulate is vacuously true because there are no lines in trivial geometry. The Protractor Postulate is vacuously true because there are no lines or rays to make angles. Finally, the Side-Angle-Side Postulate is vacuously true as well because there are no segments in trivial geometry to make triangles. None of the parallel postulates are satisfied in trivial geometry because there is only one point.
\end{proof}
\begin{problem}{4.2.1}
Prove the Converse to the Isosceles Triangle Theorem (Theorem 4.2.2).
\end{problem}
\begin{proof}{(Brianna Hillman)}
Let $\triangle {ABC}$ be a triangle such that $\angle ABC \cong \angle ACB$. We want to show that $\overline{AB} \cong \overline{AC}$. By the Existence and Uniqueness of Perpendiculars, we can drop a perpendicular from $A$ to some point $P$ on $\overline{BC}$. Then, since $\angle ABC \cong \angle ACB, AP = AP$, and $\angle APB \cong \angle APC$, $\triangle ABP \cong \triangle APC$ by AAS. Therefore, $\overline{AB} \cong \overline{AC}$.
\end{proof}
\begin{problem}{4.2.4}
Prove the HYpotenus-Leg Theorem
\end{problem}
\begin{proof}
(Katelynn Gordon)
Let $\triangle {ABC}$ be a triangle such that $\mu (\angle {BAC})$=90$^{\circ}$. Then extend $\overrightarrow{AC}$ so that $\overline{CA} \cong \overline{AF}$. Now forming $\overline{FB}$ creates two triangles $\triangle{ABC}$ and $\triangle{ABF}$. $\triangle{ABC} \cong \triangle{ABF}$ by SAS so $\overline{FB} \cong \overline{CB}$.
\end{proof}
\begin{problem}{4.2.5}
Prove that it is possible to construct a congruent copy of a triangle on a given base (Theorem 4.2.6).
\end{problem}
\begin{proof}
(Victoria Krohn) Let $\triangle ABC$ be a triangle, $\overline DE$ is a segment such that $\overline DE \cong \overline AC$, and $H$ is a half-plane bounded by $\overleftrightarrow DE$. We need to show $\triangle ABC$ is congruent to another triangle contructed from $\overleftrightarrow DE$.
Let point $G$ be in $H$. By the Angle Construction Postulate, construct $\overrightarrow DG$ such that $\angle BAC \cong \angle GDE$.
By the Point Construction Postulate, let point $F$ lie on $\overrightarrow DG$ such that $\overline AB \cong \overline DF$. Form $\triangle DFE$. By SAS, $\triangle ABC \cong \triangle DFE$ because $\overline AB \cong \overline DF$, $\angle BAC \cong \angle GDE$, and $\overline DE \cong \overline AC$.
\end{proof}
\begin {problem}{4.3.7}
Prove that the shortest distance from a point to a line is measured along the perpendicular (Theorem 4.3.4)
\end{problem}
\begin{proof}
(Katelynn Gordon) Let there be a line $\ell$ such that $F$, $R$ $\in$ $\ell$ and $F$ $\neq$ $R$. Then let $P$ be a point such that $P$ $\notin$ $\ell$. Then we can drop a perpendicular from $P$ to $\ell$ that goes through $F$. We can also connect $P$ to $R$ in order to form $\triangle$ $PFR$. We know that $\angle$ $PFR$ is a right angle since it is formed by a perpendicular. We then know that $\triangle$ $PFR$ is a right triangle and therefore we know that the hypotenuse which is across from the right angle of a right triangle in the longest side of the triangle by the Scalene Inequality. Therefore, the perpedicular is shorter than the hypotenuse of the right triangle.
\end{proof}
\begin{problem}{4.3.8}
Prove the Pointwise Characterization of Angle Bisectors (Theorem 4.3.8)\\
Let $A$ and $B$ be distinct points. A point $P$ lies on the perpendicular bisector of $\overline{AB}$ if and only if $PA = PB$.
\end{problem}
\begin{proof} (Victoria Krohn)
$\Leftarrow$ Assume $d(P,\overleftrightarrow{AB}) = d(P\overleftrightarrow{AC})$. By the Existence and Uniqueness of Perpendiculars, drop a perpendicular from $P$ to point $N$ on $\overrightarrow{AC}$ and to point $M$ on $\overrightarrow{AB}$. By the definiton of perpendicular, $\mu(\angle PNA) = 90^\circ$ and $\mu(\angle PMA) = 90^\circ$. By definiton of congruce, $\angle PNA = \angle PMA$. By the Hypotenuse - Leg Theorm, $\triangle ANP \cong \triangle AMP$ because $\overline{AP} \cong \overline{AP}$, $\angle PNA = \angle PMA$, and $\overline{NP} \cong \overline{MP}$ (by the assumptions). By definiton of congruent triangles, $\angle PAN \cong \angle PAM$, therefore $\overrightarrow{AP}$ is an angle bisector of $\angle CAB$.
$\Rightarrow$ Assume $\overrightarrow{AP}$ is an angle bisector of $\angle CAB$. Let $N$ be a point on $\overrightarrow{AC}$ and $M$ be a point on $\overrightarrow{AB}$. By the Existence and Uniqueness of Perpendiculars, drop a perpendicular from $P$ to $N$ and from $P$ to $M$. By the definiton of perpendicular, $\mu(\angle PNA) = 90^\circ$ and $\mu(\angle PMA) = 90^\circ$. By definiton of congruce, $\angle PNA \cong \angle PMA$. By Angle-Angle-Side, $\triangle ANP \cong \triangle AMP$ because $\overline{AP} \cong \overline{AP}$, $\mu(\angle PAN) = \mu(\angle PAM)$ (by definition of Angle Bisectors) so $\angle PAN \cong \angle PAM$ (by defintion of congruence), and $\angle PNA \cong \angle PMA$. Thus, by definiton of congruent triangles, $\overline{NP} \cong \overline{MP}$.
\end{proof}
\begin{problem}{4.4.3}
Prove Corollary 4.4.8 If $l$, $m$ and $n$ are three lines such that $m\perp l$ and $n\perp n$, then either $m=n$ or $m\parallel n$.
\end{problem}
\begin{proof}
(Katelynn Gordon) Assume $m \neq n$. Then we want to show that $m \parallel n$. Then know that if $l \perp m$ and $l \perp n$ then $l$ is a transversal. Then by the definition of perpendicular, we know that $l$ crosses $m$ and $n$ at 90$^{\circ}$. We can then see that have corresponding congruent angles so it follows that $m\parallel n$.
\end{proof}
\begin{problem}{4.6.6}
Prove that a quadrilateral is convex if the diagonals have a point in common (the remaining part of Thm 4.6.8).
\end{problem}
\begin{proof}
(Connor Lowman) Let $\square ABCD$ be a quadrilateral and let diagonals $\overrightarrow{AC} \cap \overrightarrow{BD}$ at interior point E. So $D \ast E \ast B$ and $A \ast E \ast C$. Then E is in the intersection of the half-plane formed by $\overleftrightarrow{DC}$ and A, and the half-plane formed by $\overleftrightarrow{AD}$ and C. Thus E is in the interior of $\angle{ADC}$. By the Ray Theorem, E and B are on the same side of $\overleftrightarrow{DC}$. Therefore, B is in the interior of $\angle{ADC}$.
\end{proof}
\begin{problem}{4.6.10}
Let $\square ABCD$ be a convex quadrilateral. Prove that each of the following conditions implies that $\square ABCD$ is a parallelogram.
\end{problem}
a. $\triangle ABC \cong \triangle CDA$
\begin{proof}
(Connor Lowman) Let $\square ABCD$ be a convex quadrilateral such that $\triangle ABC \cong \triangle CDA$. Notice that $\overleftrightarrow{AC}$ is a transversal cutting through $\overleftrightarrow{AB}$ and $\overleftrightarrow{DC}$. Since $\triangle ABC \cong \triangle CDA$, $\angle ACB \cong \angle DAC$ and $\angle DCA \cong \angle BAC$. Thus, by Alternate Interior Angles Theorem, $\overleftrightarrow{AB} \| \overleftrightarrow{DC}$. Notice $\overleftrightarrow{AC}$ is also a transversal cutting through $\overleftrightarrow{AD}$ and $\overleftrightarrow{BC}$. Again, angles $\angle ACB$,$\angle DAC$ and $\angle DCA$,$\angle BAC$ are alternate interior angles. Similarly, by Alternate Interior Angles Theorem, $\overleftrightarrow{AD} \| \overleftrightarrow{BC}$. Therefore, $\square ABCD$ is a parallelogram.
\end{proof}
b. $AB=CD$ and $BC=AD$.
\begin{proof}
(Connor Lowman) Let $\square ABCD$ be a convex quadrilateral such that $AB=CD$ and $BC=AD$. Consider diagonal $\overline{AC}$. Then by SSS, $\triangle ABC \cong \triangle CDA$. Therefore, by proof from part a, $\square ABCD$ is a parallelogram.
\end{proof}
d. The diagonals $\overline{AC}$ and $\overline{BD}$ share a common midpoint.
\begin{proof}
(Kathryn Bragwell) Let $\square ABCD$ be a convex quadrilateral. Let $E$ be a common midpoint $\overline{AC}$ and $\overline{BD}$ share. Thus $AE=EC$ and $DE=CE$. Notice $\angle{BEC}$ and $\angle{AED}$ are vertical angles thus $\angle{BEC} \cong \angle{AED}$ also $\angle{CED} \cong \angle {BED}$ because they are vertical angles also. Thus $\triangle{ABE} \cong \triangle{CDE}$ by ASA and $\triangle{CBE} \cong \triangle{AED}$ by ASA. Since $\overline{AC}$ is a transversal of $\overline{AB}$ and $\overline{DC}$, and $\angle{BAD} \cong \angle{ECD}$ then $\overline{AB} \parallel \overline{DC}$. Similarly $\overline{BC} \parallel \overline{AD}$.
\end{proof}
\noindent \hrulefill
\vspace{-1cm}
\begin{center}
End material for Exam 1 / Begin material for Exam 2.
\end{center}
\vspace{-1cm}
\hrulefill
\begin{problem}{5.1.5}
Properties of 60-60-60 and 30-60-90 triangles. An equilateral triangle is one in which all 3 sides have equal lengths.
(a) Prove that a Euclidean triangles is equilateral if and only if each of its angles measures $60^\circ$.
\begin{proof}
$\Longleftarrow$ Let $\triangle ABC$ be a triangle such that all angles are the same. By the Angle Sum Theorem $\angle ABC + \angle BCA + \angle CAB = 180, 180/3 = 60$ thus, all 3 angles are $60^\circ$. By the Converse to the Isosceles Triangle Theorem, $\mu ( \angle CAB ) = \mu ( \angle BCA )$ so $\overline{AB} \cong \overline{BC}$ and $\mu ( \angle ABC ) = \mu ( \angle CAB )$ so $\overline{BC} \cong \overline{AC}$. Therefore, $\overline{AB} \cong \overline{BC} \cong \overline{AC}$.
$\Longrightarrow$ Assume all sides of $\triangle ABC$ are congruent. By the Isosceles Triangle Theorem, base angles $\mu ( \angle CAB ) = \mu ( \angle BCA )$, because sides $\overline{AB} \cong \overline{BC}$ and base angles $\mu ( \angle ABC ) = \mu ( \angle CAB )$ because $\overline{BC} \cong \overline{AC}$. Therefore, $\mu ( \angle ABC ) = \mu ( \angle CAB ) = \mu ( \angle BCA )$.
\end{proof}
(b)Prove that there is an equilateral triangle in Euclidean geometry.
\begin{proof}
Let $A, B, C, D$ be 4 distinct points such that $\overline{AB}$ creates 2 half planes and that point $D$ is in one of the half planes, to construct $\overrightarrow{AD}$ such that $\mu (\angle DAB) = 60^\circ$ (by the Angle Construction Postulate). Let point $C$ lie on $\overrightarrow{AD}$ such that $A*C*D*$ and that $\overline{AC} \cong \overline{AB}$ (by the Point Construction Postulate). Let $\overline{CB}$ form to construct $\triangle ABC$. By the Isosceles Triangle Theorem, base angles $\mu(\angle ACB)=\mu(\angle ABC)$ because $\overline{AC} \cong \overline{AB}$. By the Angle Sum Postulate, $\angle ABC + \angle ACB + \angle CAB = 180$, $\mu(\angle ACB)=\mu(\angle ABC)$ so $2(\angle ABC) + \angle CAB = 180$ Thus, $\angle CAB = 180 - 2(\angle ABC)$, so $180/3=60^\circ$ and $\mu ( \angle ABC ) = \mu ( \angle CAB ) = \mu ( \angle ACB )$. By part a, all sides are equal thus $\triangle ABC$ is an equilateral triangle.
\end{proof}
(c) Split an equilateral triangle at the midpoint of one side to prove that there is a triangle whose angles measure $30^\circ$, $60^\circ$, and $90^\circ$.
\begin{proof}
Let $\triangle ABC$ be an equilateral triangle and let point $D$ be the midpoint
\end{proof}
\end{problem}
\begin{problem}{5.3.2}
Prove the SAS Similarity Criterion (Theorem 5.3.3).
\end{problem}
\begin{proof}
(Katelynn Gordon) Let $\triangle ABC$ with $C'$ $\in$ $\overline A \overline C$ and $\triangle DEF$ be triangles such that $\overline {AC'} \cong \overline{DF}$, $\angle{CAB} \cong \angle{FDE}$, and AB/AC=DE/DF. Now by the Incidence Postulate, form a parallel line $m$ to $\overline C \overline B$ through point $C'$. Then $m$ crosses $\overline A \overline B$ at point $B'$. Similary form a parallel line to $m$ called $l$ through point $A$. Then $\angle{BAC} \cong \angle{BAC}$ since it is the same angle, $\angle{AB'C} \cong \angle{ABC}$ by properties of parallel lines, and $\angle{ACB} \cong \angle {AC'B'}$ by parallel line properties. So $\triangle{ABC} \cong \triangle{AB'C'}$ by since all of their angles are congruent. By the Parallel Projection Postulate AB'/AB=AC'/AC.
Then AB'AC/AC=ABAC'/AC. It follows that AB'/AC'=ABAC'/AC. Then AB'/AC'=AB/AC. From the assumption, AB/AC=DE/DF=AB'/AC'. Then AB'/AC=DE/DF, and AB'=DE.
so $\triangle{AB'C'} \cong \triangle{DEF}$ by SAS and then $\triangle{ABC} \sim \triangle{DEF}$.
\end{proof}
\begin{problem}{5.4.3}
Prove the converse to the Pythagorean Theorem (Theorem 5.4.5)
\end{problem}
\begin{proof}(Kathryn Bragwell)
Let $\triangle{ABC}$ be triangle such that $a^2$+$b^2$=$c^2$. By the angle construction postulate let $\angle{DEF}$=$90^\circ$. By the ruler postulate let $EG$=$AC$. Since $AC$ is across from the angle with the vertex B it is $b$ according to notation. Thus $EG$=$AC$=$b$. Now, by the ruler postulate let $EA$=$CB$. Since $CB$ is across from the angle with the vertex $A$ it is $a$. Thus $EA$=$CB$=$a$. Thus we have a right triangle $\triangle{HEG}$. Let $GH$=$d$. So $\triangle{HEG}$ has a relationship of $a^2$+$b^2$=$d^2$ by the Pythagorean Theorem. Since $a^2$+$b^2$=$c^2$ and $a^2$+$b^2$=$d^2$, then $c^2$=$d^2$, $\sqrt{c}$=$\sqrt{d}$, $c$=$d$. Thus $\triangle{ABC}\cong\triangle{HEG}$ by SSS. Hence if $\triangle{ABC}$ is a triangle such that $a^2$+$b^2$=$c^2$, then $\angle{BCA}$ is a right angle.
\end{proof}
\begin{problem}{7.2.5}
Let $\square ABCD$ be Euclidean parallelogram. Choose one side as a base and define the corresponding height for the parallelogram. Prove that the area of the parallelogram is the length of the base times the height.
\end{problem}
\begin {proof}(Kathryn Bragwell)
Let $\square ABCD$ be a parallelogram. Drop a perpendicular from $\overline{DC}$ to $\overline{AB}$ at $D$. Let $E$ be the foot of the perpendicular. Drop another perpendicular from $\overline{AB}$ to $\overline{DC}$ at $B$. Let $F$ be the foot of the perpendicular. Thus there exist a right triangle $\triangle{DEA}$ and a right triangle $\triangle{CFB}$. Since $\overline{DE}$ and $\overline{FB}$ is perpendicular to $\overline{DF}$ and $\overline{BF}$, $\square BEFD$ is a rectangle by the definition of rectangles. By theorem 7.2.3 $\alpha\triangle{DEA}$ = 1/2(BE)(ED) and $\alpha\triangle{CFB}$ = 1/2(CF)(FB). By the Euclidean Area Postulate $\alpha\square{BEFD}$=(EB)(FB). Since $\triangle{DEA}$, $\alpha{CFB}$, and $\square{BEFD}$ exist within $\square{ABCD}$ we can add their areas. Thus $\alpha\square{ABCD}$=1/2(AE)(ED)+1/2(CF)(FB)+(EB)(FB). Since $\square{BEFD}$ is a rectangle it is also a parallelogram , therefore by properties of parallelograms we know $\overline{FB}\cong\overline{EB}$. Thus $\alpha\square {ABCD}$ =1/2(AE)(FB)+1/2(CF)(FB)+(EB)(FB). Likewise, since $\square{ABCE}$ is a parallelogram $\overline{AD}\cong\overline{BC}$. Now consider $\triangle{AED}$ and $\triangle{CFB}$, $\overline{ED}\cong\overline{FB}$, $\angle{AEB}\cong\angle{BFC}$, and $\overline{AD}\cong\overline{BC}$ therefore they are congruent by the hypotenuse leg theorem. Hence $\overline{AE}\cong\overline{FC}$. Thus $\alpha\square{ABCD}$ = 1/2 (AE)(FB)+1/2(AE)(FB)+(EB)(FB)=(AB)(FB)+(EB)(FB)=(FB)(AB+EB). Since $\overline{FB}$ is one of the perpendiculars we originally droped it is the height of $\square{ABCD}$ by the definition of height. AE+EB is the length of $\overline{AB}$ this is the base by the definition of base. Thus the area of $\square{ABCD}$ is base times height,
\end{proof}
\end{document}$/