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 𝐴𝐵,
there is a unique line that is perpendicular to line 𝐴𝐵 that passes through a
given point 𝐶 on line 𝐴𝐵. 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 𝐶. 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 𝐴𝐵 𝐵 prime and 𝐴 prime, respectively. We note that 𝐵 prime 𝐶 equals 𝐴
prime 𝐶, since these are both radii of the same circle centered on point 𝐶. So we can say that 𝐶 is
equidistant from 𝐴 prime and 𝐵 prime.
Next, we will set the radius of our
compass to be larger than 𝐴 prime 𝐶 and construct a pair of congruent circles
centered at 𝐵 prime and 𝐴 prime. Instead of constructing full
circles, we can just sketch the arcs where the circles intersect, either above or
below line 𝐴𝐵. In this case, we call the point of
intersection above line 𝐴𝐵 point 𝐷.
We note that 𝐵 prime 𝐷 equals 𝐴
prime 𝐷, as they are radii of congruent circles, one centered on point 𝐵 prime and
the other centered on point 𝐴 prime. Finally, we use our straightedge to
connect points 𝐶 and 𝐷 with a line segment. By the reflexive property, 𝐷𝐶
equals 𝐷𝐶. We note that triangles 𝐵 prime
𝐶𝐷 and 𝐴 prime 𝐶𝐷 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 𝐴 prime 𝐶𝐷 is
congruent to angle 𝐵 prime 𝐶𝐷.
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 𝐶𝐷 is a unique
line perpendicular to line 𝐴𝐵 passing through point 𝐶.
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
𝐴𝐵 passing through point 𝐶 on line 𝐴𝐵, we use the following steps. We start by setting the point of
our compass at 𝐶, then tracing a circle intersecting line 𝐴𝐵 twice. We label these points 𝐵 prime and
𝐴 prime. Then, we set the radius of the
compass to be larger than 𝐴 prime 𝐶, and we trace arcs of congruent circles
centered at 𝐵 prime and 𝐴 prime such that they intersect. We call the point of intersection
𝐷. Finally, we have that line 𝐶𝐷 is
perpendicular to line 𝐴𝐵. 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
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 𝐴𝐵, we use the following steps. We begin by setting the radius of
the compass to be greater than half the length of 𝐴𝐵. Then, we trace two arcs of the
circle of this radius centered at 𝐴 and two intersecting arcs of a congruent circle
centered at 𝐵. These arcs will intersect at two
points that we label 𝐶 and 𝐷.
We then connect 𝐶 and 𝐷 with a
straight line. We claim that line 𝐶𝐷 is the
perpendicular bisector of line segment 𝐴𝐵. 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 𝐴 and 𝐵. Because the radii are equal in
length, it follows that triangle 𝐵𝐶𝐷 is congruent to triangle 𝐴𝐶𝐷 by the
side-side-side triangle congruence criterion. We know that their corresponding
angles must be congruent as well, so we have angle 𝐵𝐶𝐷 congruent to angle
We label the intersection of 𝐶𝐷
and 𝐴𝐵 point 𝐸. We see that triangle 𝐴𝐸𝐶 and
triangle 𝐵𝐸𝐶 share two congruent sides and the included angles are congruent. Thus, triangle 𝐴𝐸𝐶 is congruent
to triangle 𝐵𝐸𝐶 by the side-angle-side triangle congruence criterion.
We know that their corresponding
sides must be congruent, so we have 𝐴𝐸 equal to 𝐸𝐵. We also note that angle 𝐴𝐸𝐶 is
congruent to angle 𝐵𝐸𝐶. Since these angles sum to make a
straight angle, they must be right angles.
In summary, we have just shown that
line 𝐶𝐷 separates line segment 𝐴𝐵 into two equal parts and meets 𝐴𝐵 at right
angles. This means that line 𝐶𝐷 is the
perpendicular bisector of line segment 𝐴𝐵.
Now let’s see an example of
determining which quadrilateral is given by a geometric construction.
Draw the given figure and
connect the points 𝐴𝐶𝐵𝐷. What does that figure
Before connecting 𝐴𝐶𝐵𝐷,
let’s label the intersection of line segments 𝐶𝐷 and 𝐴𝐵 point 𝐸. We see from the diagram that
𝐴𝐸 and 𝐸𝐵 have the same length and 𝐶𝐸 and 𝐸𝐷 have the same length. So we know that 𝐶𝐷 bisects
line segment 𝐴𝐵 and 𝐴𝐵 bisects line segment 𝐶𝐷. We will now use a straightedge
to connect the points 𝐴, 𝐶, 𝐵, and 𝐷 to form quadrilateral 𝐴𝐶𝐵𝐷.
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 𝐴 with a radius 𝐴𝐶, and the pink circle is centered at 𝐵 with radius
𝐵𝐶. We recognize this construction
as the perpendicular bisector construction.
This means that line 𝐶𝐷 not
only bisects line segment 𝐴𝐵, but it is the perpendicular bisector of line
segment 𝐴𝐵. Therefore, 𝐶𝐵 bisects 𝐴𝐵 at
right angles, which we can add to the diagram. Therefore, the diagonals of
𝐴𝐶𝐵𝐷 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 𝐴𝐸𝐶, 𝐵𝐸𝐶, 𝐵𝐸𝐷, and 𝐴𝐸𝐷 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 𝐴𝐶𝐵𝐷 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 𝐴𝐸
and 𝐶𝐸 have the same length, we can’t prove that the diagonals are
congruent. And therefore, we can’t claim
that 𝐴𝐶𝐵𝐷 is a square.
Based on the evidence of the
perpendicular diagonals and the congruent sides, we conclude that quadrilateral
𝐴𝐶𝐵𝐷 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 𝐴𝐵 through a point 𝐶 on the line. But what if 𝐶 does not lie on the
line? To find the perpendicular to line
𝐴𝐵 passing through a point 𝐶 not on the line, we use the following steps. First, we place the point of our
compass on 𝐶. Then we trace the arc of a circle
that intersects line 𝐴𝐵 twice. We label these points 𝐵 prime and
𝐴 prime. We note that 𝐶 is equidistant from
𝐵 prime and 𝐴 prime, because 𝐶𝐵 prime and 𝐶𝐴 prime are radii of the same
circle centered at 𝐶.
Next, we want to find another point
equidistant from 𝐵 prime and 𝐴 prime below the line 𝐴𝐵. We do this by tracing the arcs of
two intersecting circles centered at 𝐵 prime and 𝐴 prime using the same radius we
started with. We name the point of intersection
𝐷. Since we kept the radii of the
circles the same, 𝐵 prime 𝐷 equals 𝐴 prime 𝐷 equals 𝐵 prime 𝐶 equals 𝐴 prime
𝐶, which we add to the diagram. We recognize the resulting
quadrilateral as a rhombus.
Next, we use our straightedge to
connect vertex 𝐶 to vertex 𝐷 in the diagram. We recall that the diagonals of a
rhombus are perpendicular. Therefore, line segment 𝐶𝐷 is
perpendicular to line 𝐴𝐵. We can follow these same steps to
construct the perpendicular line through any point not on the line. Let’s 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 𝐶 is
not on line 𝐴𝐵. We also see a circle centered
on 𝐶 intersecting the line at points 𝐷 and 𝐹. Then, we notice the
intersecting arcs of two circles centered on 𝐷 and 𝐹. This point of intersection then
connects back to point 𝐶, forming a line perpendicular to line 𝐴𝐵.
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, we’ll
begin with a point and a line. Let’s the name the point 𝐶 and
the line 𝐴𝐵. The statement provided is
regarding the shortest distance from 𝐶 to line 𝐴𝐵. The claim is that any line
segment through the point to the line is the shortest.
So let’s sketch an arbitrary
line segment from the point to the line. Let’s name the point of
intersection with the line point 𝐷. If the given statement is true,
then we will not find any shorter distance from 𝐶 to line 𝐴𝐵. Let’s try sketching the
perpendicular distance from 𝐶 to the line. We label the point 𝐸 where the
perpendicular segment meets line 𝐴𝐵 at a right angle.
We notice that we have formed a
right triangle: triangle 𝐶𝐸𝐷. Line segment 𝐶𝐷 is the
hypotenuse of this right triangle. We recall that the hypotenuse
of a right triangle is always the longest side. Therefore, 𝐶𝐷 is greater than
𝐶𝐸. This means that the given
statement is false. Any nonperpendicular line
segment from point 𝐶 to line 𝐴𝐵 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.
Let’s 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 𝐶 and label the points of intersection 𝐵 prime and 𝐴 prime. Then, we set a radius longer than
𝐴 prime 𝐶 and trace two intersecting circles centered on 𝐵 prime and 𝐴
prime. We label this intersection point
𝐷. The line passing through 𝐶 and 𝐷
is perpendicular to line 𝐴𝐵.
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 𝐶, intersecting line 𝐴𝐵 at two points, which we name 𝐵 prime
and 𝐴 prime. Next, we trace two circles with the
same radius centered at 𝐵 prime and 𝐴 prime. We label their intersection below
the line point 𝐷. Then, the line passing through
points 𝐶 and 𝐷 is perpendicular to line 𝐴𝐵.
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 𝐴𝐵. Then, we trace two circles of this
radius centered at 𝐴 and 𝐵, intersecting both above and below the line segment
𝐴𝐵. We label these intersections 𝐶 and
𝐷. The line passing through 𝐶 and 𝐷
is the perpendicular bisector of line segment 𝐴𝐵. We note that all points on the
perpendicular bisector are equidistant from both 𝐴 and 𝐵, 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.