Video Transcript
Draw triangle 𝐴𝐵𝐶 where length
𝐴𝐵 equals three centimeters, length 𝐴𝐶 equals four centimeters, and length 𝐵𝐶
equals five centimeters. Then, draw ray 𝐶𝐷, ray 𝐴𝐸, and
ray 𝐵𝐹 to bisect angles 𝐴𝐶𝐵, 𝐵𝐴𝐶, and 𝐴𝐵𝐶, respectively, where ray 𝐶𝐷
cuts line segment 𝐴𝐵 at 𝐷, ray 𝐴𝐸 cuts line segment 𝐵𝐶 at 𝐸, and ray 𝐵𝐹
cuts line segment 𝐴𝐶 at 𝐹. From your observation, which of the
following statements is true? Option (A) ray 𝐶𝐷 is
perpendicular to line segment 𝐴𝐵. Option (B) ray 𝐵𝐹 is
perpendicular to line segment 𝐴𝐶. Option (C) ray 𝐴𝐸 is
perpendicular to line segment 𝐵𝐶. Option (D) rays 𝐶𝐷, 𝐴𝐸, and
𝐵𝐹 intersect at one point. Or option (E) all of the answers
are correct.
We are asked to draw a triangle
with specific side lengths in centimeters and then construct rays that bisect each
interior angle and intercept the opposite side of the triangle. To do this, we will need a ruler
and a compass. Once we have the diagram drawn, we
will consider each of the possible answers. To begin, we use a ruler to measure
out each side of the triangle so that 𝐴𝐵 equals three centimeters, 𝐴𝐶 equals
four centimeters, and 𝐵𝐶 equals five centimeters. We note that together lengths
three, four, and five are considered a Pythagorean triple and thus will always form
a right triangle. So, we know that side 𝐴𝐵 is
perpendicular to side 𝐴𝐶.
Next, we will use a compass and
ruler to construct the bisector of angle 𝐴𝐵𝐶. We recall the first step is to
trace a circle using the compass centered at the vertex and intersecting both sides
of the angle. We will need to do the same for
angles 𝐵𝐴𝐶 and 𝐴𝐵𝐶 as well. The next step to bisect angle
𝐴𝐵𝐶 is to construct two circles of the same radius centered at the points of
intersection constructed in the first step. These two circles intersect, as
shown. Then, we use a ruler to draw the
angle bisector from vertex 𝐵 through this point of intersection, like so. We have just drawn ray 𝐵𝐹, which
cuts side 𝐴𝐶 at 𝐹. So we label the intersection point
𝐹.
We then repeat this process to
bisect angle 𝐴𝐶𝐵, which intersects the opposite side at point 𝐷. And finally we bisect angle 𝐵𝐴𝐶,
which intersects the opposite side at point 𝐸. This is all done with only a ruler
and a compass.
From our observation, we see that
the three angle bisectors intersect at one point. Let’s call this point of
concurrency 𝑃. This confirms that answer (D) is
correct. However, answer (E) prompts us to
consider whether the first three answers might also be correct. Answers (A), (B), and (C) require
each angle bisector to intersect its opposite side at a 90-degree angle. It is clear from our carefully
drawn diagram that none of the angle bisectors are perpendicular to the sides. Therefore, it is not possible for
all the answers here to be correct.
From our observation, only answer
(D) is correct. In fact, under these conditions,
this property always holds true. The internal angle bisectors of any
triangle intersect at a single point of concurrency.
