### Video Transcript

In this video, we will learn how to
construct, using a ruler and a pair of compasses, the perpendicular to a given line
from or at a given point and the perpendicular bisector of a line segment.

We note that for any line π΄π΅,
there is a unique line that is perpendicular to line π΄π΅ that passes through a
given point πΆ on line π΄π΅. We can construct this line with a
compass and a straightedge by using our knowledge of circles and triangles. We start by setting the point of
our compass at πΆ. Then, we can use any size for the
radius. There is no need to trace the whole
circle. So we only show the arcs where the
circle intersects the line.

We call the points of intersection
of the circle and line π΄π΅ π΅ prime and π΄ prime, respectively. We note that π΅ prime πΆ equals π΄
prime πΆ, since these are both radii of the same circle centered on point πΆ. So we can say that πΆ is
equidistant from π΄ prime and π΅ prime.

Next, we will set the radius of our
compass to be larger than π΄ prime πΆ and construct a pair of congruent circles
centered at π΅ prime and π΄ prime. Instead of constructing full
circles, we can just sketch the arcs where the circles intersect, either above or
below line π΄π΅. In this case, we call the point of
intersection above line π΄π΅ point π·.

We note that π΅ prime π· equals π΄
prime π·, as they are radii of congruent circles, one centered on point π΅ prime and
the other centered on point π΄ prime. Finally, we use our straightedge to
connect points πΆ and π· with a line segment. By the reflexive property, π·πΆ
equals π·πΆ. We note that triangles π΅ prime
πΆπ· and π΄ prime πΆπ· have congruent sides. So by the side-side-side triangle
congruence criterion, these are congruent triangles. Then, because corresponding parts
of congruent triangles are congruent, it follows that angle π΄ prime πΆπ· is
congruent to angle π΅ prime πΆπ·.

For two congruent angles to make up
a straight angle, as we see here, their angles must sum to 180 degrees. And in this case, since the two
angles are congruent, they must both be right angles. This proves that πΆπ· is a unique
line perpendicular to line π΄π΅ passing through point πΆ.

Now we will clear a bit of space in
order to summarize these steps we used to construct the perpendicular.

To find the perpendicular line to
π΄π΅ passing through point πΆ on line π΄π΅, we use the following steps. We start by setting the point of
our compass at πΆ, then tracing a circle intersecting line π΄π΅ twice. We label these points π΅ prime and
π΄ prime. Then, we set the radius of the
compass to be larger than π΄ prime πΆ, and we trace arcs of congruent circles
centered at π΅ prime and π΄ prime such that they intersect. We call the point of intersection
π·. Finally, we have that line πΆπ· is
perpendicular to line π΄π΅. This technique allows us to find
the perpendicular line passing through any given point on a line.

Next, we will look at a special
case of this that splits a segment into two equal parts called the perpendicular
bisector.

In general, to bisect a line
segment is to separate it into two equal parts, in other words to cross through the
midpoint of the line segment. In particular, the perpendicular
bisector of a line segment is the line that separates the segment into two equal
parts and meets the segment at right angles. We will use our compass and
straightedge again to construct the perpendicular bisector of any line segment.

To find the perpendicular bisector
to line segment π΄π΅, we use the following steps. We begin by setting the radius of
the compass to be greater than half the length of π΄π΅. Then, we trace two arcs of the
circle of this radius centered at π΄ and two intersecting arcs of a congruent circle
centered at π΅. These arcs will intersect at two
points that we label πΆ and π·.

We then connect πΆ and π· with a
straight line. We claim that line πΆπ· is the
perpendicular bisector of line segment π΄π΅. But we still need to prove this is
true. To show our reasoning, we use our
straightedge to draw in the radii of the circles centered at π΄ and π΅. Because the radii are equal in
length, it follows that triangle π΅πΆπ· is congruent to triangle π΄πΆπ· by the
side-side-side triangle congruence criterion. We know that their corresponding
angles must be congruent as well, so we have angle π΅πΆπ· congruent to angle
π΄πΆπ·.

We label the intersection of πΆπ·
and π΄π΅ point πΈ. We see that triangle π΄πΈπΆ and
triangle π΅πΈπΆ share two congruent sides and the included angles are congruent. Thus, triangle π΄πΈπΆ is congruent
to triangle π΅πΈπΆ by the side-angle-side triangle congruence criterion.

We know that their corresponding
sides must be congruent, so we have π΄πΈ equal to πΈπ΅. We also note that angle π΄πΈπΆ is
congruent to angle π΅πΈπΆ. Since these angles sum to make a
straight angle, they must be right angles.

In summary, we have just shown that
line πΆπ· separates line segment π΄π΅ into two equal parts and meets π΄π΅ at right
angles. This means that line πΆπ· is the
perpendicular bisector of line segment π΄π΅.

Now letβs see an example of
determining which quadrilateral is given by a geometric construction.

Draw the given figure and
connect the points π΄πΆπ΅π·. What does that figure
represent?

Before connecting π΄πΆπ΅π·,
letβs label the intersection of line segments πΆπ· and π΄π΅ point πΈ. We see from the diagram that
π΄πΈ and πΈπ΅ have the same length and πΆπΈ and πΈπ· have the same length. So we know that πΆπ· bisects
line segment π΄π΅ and π΄π΅ bisects line segment πΆπ·. We will now use a straightedge
to connect the points π΄, πΆ, π΅, and π· to form quadrilateral π΄πΆπ΅π·.

To answer this question, we
must look for key features of the quadrilateral to determine what kind of
quadrilateral we are looking at.

We recall that a quadrilateral
can be classified as a trapezoid or a kite or a parallelogram, such as a
rectangle, rhombus, or square. We note that the intersecting
arcs in the diagram represent the intersection between two circles. The orange circle is centered
at π΄ with a radius π΄πΆ, and the pink circle is centered at π΅ with radius
π΅πΆ. We recognize this construction
as the perpendicular bisector construction.

This means that line πΆπ· not
only bisects line segment π΄π΅, but it is the perpendicular bisector of line
segment π΄π΅. Therefore, πΆπ΅ bisects π΄π΅ at
right angles, which we can add to the diagram. Therefore, the diagonals of
π΄πΆπ΅π· are perpendicular. We know that the only
quadrilaterals that have perpendicular diagonals are kites, rhombuses, and
squares. Therefore, it is safe to
eliminate trapezoids and rectangles from our list of possibilities.

Based on the accumulated
evidence, it follows that triangles π΄πΈπΆ, π΅πΈπΆ, π΅πΈπ·, and π΄πΈπ· are
congruent by the side-angle-side triangle congruence criterion, since all four
triangles have two congruent sides and the included angles are all right
angles. Thus, the corresponding sides
of the triangles are also congruent. This means all four sides of
quadrilateral π΄πΆπ΅π· are equal in length.

We know that rhombuses have
four congruent sides. And because a square is a
specific type of rhombus, squares also have the same property. We can eliminate kites as a
possibility, because they have two pairs of congruent sides. We recall that for a rhombus to
be a square, the diagonals must be congruent. Since we do not know if π΄πΈ
and πΆπΈ have the same length, we canβt prove that the diagonals are
congruent. And therefore, we canβt claim
that π΄πΆπ΅π· is a square.

Based on the evidence of the
perpendicular diagonals and the congruent sides, we conclude that quadrilateral
π΄πΆπ΅π· is a rhombus.

Before we move on to the next
example, there is one more construction we need to consider.

We have already seen how to find
the perpendicular line to π΄π΅ through a point πΆ on the line. But what if πΆ does not lie on the
line? To find the perpendicular to line
π΄π΅ passing through a point πΆ not on the line, we use the following steps. First, we place the point of our
compass on πΆ. Then we trace the arc of a circle
that intersects line π΄π΅ twice. We label these points π΅ prime and
π΄ prime. We note that πΆ is equidistant from
π΅ prime and π΄ prime, because πΆπ΅ prime and πΆπ΄ prime are radii of the same
circle centered at πΆ.

Next, we want to find another point
equidistant from π΅ prime and π΄ prime below the line π΄π΅. We do this by tracing the arcs of
two intersecting circles centered at π΅ prime and π΄ prime using the same radius we
started with. We name the point of intersection
π·. Since we kept the radii of the
circles the same, π΅ prime π· equals π΄ prime π· equals π΅ prime πΆ equals π΄ prime
πΆ, which we add to the diagram. We recognize the resulting
quadrilateral as a rhombus.

Next, we use our straightedge to
connect vertex πΆ to vertex π· in the diagram. We recall that the diagonals of a
rhombus are perpendicular. Therefore, line segment πΆπ· is
perpendicular to line π΄π΅. We can follow these same steps to
construct the perpendicular line through any point not on the line. Letβs see if we can recognize one
of these new constructions in our next example.

Which of the following
constructions represents drawing a perpendicular from a point lying outside a
straight line? We recall that to find a
perpendicular from a point not on a line, we first trace a circle centered at
that point which will intersect the line at the two distinct points. Then, we would trace circles
with the same radius centered on the two points of intersection. These two circles would
intersect at a point which we connect back to the original point, resulting in a
line segment perpendicular to the original line.

Of the five diagrams given, the
only one that matches these construction steps is option (E). First, we notice point πΆ is
not on line π΄π΅. We also see a circle centered
on πΆ intersecting the line at points π· and πΉ. Then, we notice the
intersecting arcs of two circles centered on π· and πΉ. This point of intersection then
connects back to point πΆ, forming a line perpendicular to line π΄π΅.

It is worth noting that the
other options give different geometric constructions. Option (A) is the construction
of a perpendicular bisector. Option (B) finds the
perpendicular to a straight line passing through a point on the line. Option (C) is a bisector;
however, it is not perpendicular. And option (D) is the
construction of an angle bisector.

In our final example, we will
consider how to find the shortest distance between a point and a line.

True or False: The shortest
distance from a point to a line equals the length of any line segment that
passes through the point and intersects with the line.

To answer this question, weβll
begin with a point and a line. Letβs the name the point πΆ and
the line π΄π΅. The statement provided is
regarding the shortest distance from πΆ to line π΄π΅. The claim is that any line
segment through the point to the line is the shortest.

So letβs sketch an arbitrary
line segment from the point to the line. Letβs name the point of
intersection with the line point π·. If the given statement is true,
then we will not find any shorter distance from πΆ to line π΄π΅. Letβs try sketching the
perpendicular distance from πΆ to the line. We label the point πΈ where the
perpendicular segment meets line π΄π΅ at a right angle.

We notice that we have formed a
right triangle: triangle πΆπΈπ·. Line segment πΆπ· is the
hypotenuse of this right triangle. We recall that the hypotenuse
of a right triangle is always the longest side. Therefore, πΆπ· is greater than
πΆπΈ. This means that the given
statement is false. Any nonperpendicular line
segment from point πΆ to line π΄π΅ would be considered a hypotenuse and thus be
longer than the perpendicular distance.

We conclude that the
perpendicular distance between a point and a line is the shortest distance.

Letβs finish by recapping some of
the important points from this video.

We learned how to construct a
perpendicular line through a point on the line. The first step is to trace a circle
centered on point πΆ and label the points of intersection π΅ prime and π΄ prime. Then, we set a radius longer than
π΄ prime πΆ and trace two intersecting circles centered on π΅ prime and π΄
prime. We label this intersection point
π·. The line passing through πΆ and π·
is perpendicular to line π΄π΅.

We also learned how to find the
perpendicular to a line through a point not on the line. We begin by tracing a circle
centered on point πΆ, intersecting line π΄π΅ at two points, which we name π΅ prime
and π΄ prime. Next, we trace two circles with the
same radius centered at π΅ prime and π΄ prime. We label their intersection below
the line point π·. Then, the line passing through
points πΆ and π· is perpendicular to line π΄π΅.

We also learned how to construct
the perpendicular bisector of a line segment. A perpendicular bisector separates
the line into two equal parts and meets the line segment at right angles. To begin, we set the radius of the
compass to be greater than half the length of π΄π΅. Then, we trace two circles of this
radius centered at π΄ and π΅, intersecting both above and below the line segment
π΄π΅. We label these intersections πΆ and
π·. The line passing through πΆ and π·
is the perpendicular bisector of line segment π΄π΅. We note that all points on the
perpendicular bisector are equidistant from both π΄ and π΅, since the distance is
given by the same radius length.

And finally, we learned that the
shortest distance between a line and a point is the perpendicular distance.