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Instructions. You will submit your solutions on Gradescope, which you can access
through our Canvas course page by clicking on the assignment.
You are welcome to submit your homework either handwritten or in LaTeX. If you
choose to submit handwritten work, please use pen rather than pencil, and make sure
that both your handwriting and your scans are as readable as possible in order to
facilitate grading.
Please also feel free to attend office hours if you would like hints on the homework or
if you would like to discuss more advanced material. Also, you are highly encouraged
to work in groups!
You will see that our Ximera course page proposes you to see the answers, but this
will only become available once I publish the solution.
Parallelograms (45 points total).
A quadrilateral is a figure determined by four distinct points \(A,B,C,D\) such that no three lie
on the same straight line, and the segments \(AB\), \(BC\), \(CD\), and \(DA\) intersect only at their
endpoints.
A quadrilateral \(ABCD\) is called a parallelogram if its opposite sides are parallel, that is,
Hint: For these exercises, you should always start with a diagram. Then, you may
rely on:
the congruence theorems I.4 (SAS), I.8 (SSS), and I.26 (AAS/ASA),
the angle equality I.15 (vertical angles),
the angle properties of parallel lines I.27, I.28, and I.29.
How we will grade Problem 1. We will give some credit for a correct initial
diagram, some credit for the method, some credit for the correctness of the argument,
and some credit for proving the desired conclusion. In particular, we will look at how
well you use the information given in the problem together with postulates and
previously proved propositions to deduce new consequences.
If you use a proposition, make sure to explain why its hypotheses are satisfied before
applying it.
(a)
(15 points) Show that if \(ABCD\) is a parallelogram, then
Since \(ABCD\) is a parallelogram, we have \((AB)\parallel (CD)\) and \((BC)\parallel (AD)\). Because the line \(AC\) meets the
parallel lines \((AB)\) and \((CD)\), Proposition I.29 gives
\[ \angle BAC = \angle DCA. \]
Likewise, since the line \(AC\) meets
the parallel lines \((BC)\) and \((AD)\), Proposition I.29 gives
\[ \angle BCA = \angle DAC. \]
Also, the side \(AC\) is common
to triangles \(BAC\) and \(DCA\).
Therefore triangles \(BAC\) and \(DCA\) have two equal angles and the included side equal,
so by Proposition I.26 (ASA) they are congruent. Hence the corresponding
sides are equal:
\[ AB = CD \qquad \text {and}\qquad BC = AD. \]
This is what we wanted to prove.
(b)
(15 points) Let \(ABCD\) be a parallelogram. Show that the diagonals \(AC\) and \(BD\) intersect at
their midpoint.
You may assume that \(AC\) and \(BD\) intersect at a point \(E\), and you need to show
Let \(E\) be the intersection point of the diagonals \(AC\) and \(BD\).
Consider triangles \(AEB\) and \(CED\). Since the diagonals intersect, the angles \(\angle AEB\) and \(\angle CED\)
are vertical angles, so by Proposition I.15,
\[ \angle AEB = \angle CED. \]
Since \((AB)\parallel (CD)\) and the line \(BD\) meets them, Proposition I.29 gives
\[ \angle ABE = \angle CDE. \]
From part (a),
opposite sides of a parallelogram are equal, so
\[ AB = CD. \]
Therefore triangles \(AEB\) and \(CED\) have two equal angles and one corresponding
side equal, so by Proposition I.26 (AAS) they are congruent. Hence
corresponding sides are equal:
\[ AE = EC \qquad \text {and}\qquad BE = ED. \]
Thus the diagonals bisect one another, so
they intersect at their midpoint.
(c)
(15 points) (Converse) Let \(ABCD\) be a quadrilateral such that
Now the line \(AC\) meets the lines \((AB)\) and \((CD)\), and the alternate interior angles \(\angle BAC\) and
\(\angle DCA\) are equal. Therefore, by Proposition I.27, the lines \((AB)\) and \((CD)\) are parallel.
Similarly, the line \(AC\) meets the lines \((BC)\) and \((AD)\), and the alternate interior angles
\(\angle BCA\) and \(\angle DAC\) are equal. Therefore, again by Proposition I.27,
\[ (BC)\parallel (AD). \]
Since both pairs of opposite sides are parallel, \(ABCD\) is a parallelogram.
Important. In these propositions, Euclid uses the word “equal” to refer to area. Two
figures are said to be equal if they can be subdivided into smaller pieces which are
themselves equal.
Choose three propositions among
\[ \text {I.34, I.35, I.37, I.41, I.46, I.47}, \]
and for each of your choices, draw a summary
diagram that helps you remember the statement.
How we will grade Problem 2. We will give full credit for a clean diagram with
clear markings introducing the relevant notations, together with a brief piece of text
stating the hypotheses and the conclusion, in the style used during Week 2
Lecture 1.
Remark: You are encouraged to be creative. For example, you may use different
colors (or shading) to highlight the different regions in each figure. If you do not
want to use color, you may instead distinguish the regions by using different
patterns.
For instance, Proposition I.34 asserts that in a parallelogram \(ABCD\), the triangles \(ABC\) and \(CDA\) are
equal, in the sense that they have the same area. You might highlight the region \(ABC\) in
blue and the region \(CDA\) in red, and then summarize the proposition by writing
something like
If you prefer not to use color, you could instead mark one region with one pattern
and the other region with a different pattern, and write the same kind of
conclusion.
Solution:
Since this problem asks for diagrams, below I write the short text that can
accompany three summary diagrams.
Choice 1: Proposition I.34.
Hypotheses:\(ABCD\) is a parallelogram, and the diagonal \(AC\) is drawn.
Conclusion: The diagonal cuts the parallelogram into two equal triangles:
In the diagram, one can shade \(\triangle ABC\) one way and \(\triangle CDA\) another way. GeoGebra Link
Choice 2: Proposition I.41.
Hypotheses: A parallelogram and a triangle stand on the same base and lie
between the same parallels.
Conclusion: The parallelogram has twice the area of the triangle:
\[ \text {area of parallelogram} = 2\cdot \text {area of triangle}. \]
In the diagram, one can mark the whole parallelogram with one pattern and
the triangle with another, and then write that the parallelogram is double the
triangle. GeoGebra Link
Choice 3: Proposition I.47.
Hypotheses:\(ABC\) is a right triangle with right angle at \(A\), and squares are constructed
on the three sides.
Conclusion: The square on the side opposite the right angle equals the sum of
the squares on the two legs:
\[ \text {square on }BC = \text {square on }AB + \text {square on }AC. \]
In the diagram, one can draw the three squares and mark the two smaller ones
with different patterns, indicating that together they equal the large square.
GeoGebra Link
Straight-edge and compass constructions (30 points total).
There will be a compass-and-straightedge construction problem on the midterm, to
be done by hand (no GeoGebra).
One thing you may have noticed is that Euclid does not hesitate to reuse
constructions that were established earlier. This is perfectly natural, but some of
these previously established constructions require many steps. There is another
kind of exercise, closer in spirit to a puzzle, which asks you to carry out
a construction in as few moves as possible using only a straightedge and
compass.
We first need to explain how moves are counted.
Drawing a new line counts as one move.
Drawing a new circle counts as one move.
Extending a line already drawn on the paper does not count as a move.
Choosing points at random on the paper does not count as a move.
Naming intersection points does not count as a move.
Thus, only actual uses of the straightedge to draw a new line and actual uses of the
compass to draw a new circle count as moves.
The goal in the following exercises is to complete each construction in as few moves
as possible. You will still receive full credit if your construction is correct but does
not use the minimum possible number of moves.
We will sometimes give a par value for a construction. This is the typical number of
moves an experienced constructor might use. By thinking harder, you may sometimes
do better.
For example, the construction of an equilateral triangle on a given segment \(AB\) takes \(4\)
moves in this counting system:
(a)
draw the circle with center \(A\) and radius \(AB\);
(b)
draw the circle with center \(B\) and radius \(BA\);
(c)
let \(C\) be an intersection point of the two circles, and draw the segment \(AC\);
(d)
draw the segment \(BC\).
Then \(ABC\) is the required triangle.
In particular, the segment \(AB\) is considered to be already given: you should draw it on
the paper, but you should not count it as a move.
How we will grade Problem 3. We will give points for the following:
all the lines and circles used in the construction are shown;
the diagram has clear markings;
the number of moves is stated;
a short proof is provided;
the construction achieves the required goal.
Using this convention, do the following constructions in as few moves as
possible.
(a)
(10 points) Given an angle \(\angle BAC\), construct its bisector. (par = 4)
Draw a circle centered at \(A\) meeting the rays \(AB\) and \(AC\) at points \(D\) and \(E\).
(ii)
Draw a circle centered at \(D\) with radius \(DE\).
(iii)
Draw a circle centered at \(E\) with radius \(ED\).
(iv)
Let \(F\) be an intersection point of the two circles from steps 2 and 3.
Draw the line \(AF\).
Number of moves:\(4\).
Proof. Since \(AD=AE\) (they are radii of the circle centered at \(A\)), and since \(DF=EF\)
(they are radii of the circles centered at \(D\) and \(E\)), and since \(AF\) is common,
triangles \(ADF\) and \(AEF\) are congruent by Proposition I.8. Therefore
\[ \angle DAF = \angle FAE. \]
But \(D\) lies on
the ray \(AB\) and \(E\) lies on the ray \(AC\), so
\[ \angle BAF = \angle FAC. \]
Hence the line \(AF\) bisects the angle \(\angle BAC\).
(b)
(10 points) Given a segment \(AB\), construct its midpoint. (par = 3)
Draw the circle centered at \(A\) with radius \(AB\).
(ii)
Draw the circle centered at \(B\) with radius \(BA\).
(iii)
Let \(C\) and \(D\) be the two intersection points of these circles. Draw the
line \(CD\), and let \(M\) be the intersection point of \(CD\) with \(AB\).
Number of moves:\(3\).
Proof. Because \(C\) and \(D\) lie on circles with equal radii centered at \(A\) and \(B\), we have
\[ AC=CB=BD=DA. \]
In particular, \(ACBD\) is a parallelogram thanks to Problem 1 (c). Thus, by Problem 1
(b), its diagonals intersect at their midpoint. Hence, \(E\) is the midpoint of \(AB\).
(c)
(10 points) Given a line \(\ell \) and a point \(A\) on \(\ell \), construct the line perpendicular to \(\ell \)
through \(A\). (par = 4; possibly 3 according to Hartshorne)
Draw a circle centered at \(A\) meeting the line \(\ell \) at two points \(B\) and \(C\).
(ii)
Draw a circle centered at \(B\) with radius \(BC\).
(iii)
Draw a circle centered at \(C\) with radius \(CB\).
(iv)
Let \(D\) be an intersection point of the circles from steps 2 and 3. Draw
the line \(AD\).
Number of moves:\(4\).
Proof. By Construction, \(A\) is the midpoint of \(BC\), so \(AB=BC\). Then, \(D\) is at the intersection
of two circles with equal radii, so \(CD = BD\). Finally, the side \(AD\) is common to both \(BAD\)
and \(CAD\). Thus, by the SSS Theorem, \(\angle BAD = \angle CAD\), and, since \(A\), \(B\), and \(C\) lie on the same
line, \(\angle BAD \) and \( \angle CAD\) are right angles. Hence, since the line \((AD)\) is perpendicular to \(\ell \).
Reflection (2 bonus points).
Reflect on the first three weeks of the course.
(a)
What did you find most challenging so far?
(b)
How did you find the pace of the lectures: too fast, too slow, or just right?
(c)
Is there something you wish we did more of, or less of?
Please answer in a few complete sentences. Your responses will be anonymous.
