Video Transcript
Draw triangle 𝐴𝐵𝐶 where 𝐴𝐵
equals three centimeters, 𝐵𝐶 equals four centimeters, and 𝐴𝐶 equals five
centimeters. Point 𝐷 lies on the ray from 𝐵
through 𝐶 such that 𝐷 is not on line segment 𝐵𝐶. Draw angle 𝐷𝐶𝐸 congruent to
angle 𝐶𝐵𝐴, where 𝐸 is on the upper side of line segment 𝐵𝐶. Which of the following is true? Option (A) the measure of angle 𝐴
equals the measure of angle 𝐸𝐶𝐴. Option (B) the measure of angle 𝐴
is equal to the measure of angle 𝐵𝐶𝐴. Option (C), the measure of angle 𝐴
equals the measure of angle 𝐸𝐶𝐷. Or is it option (D) the measure of
angle 𝐵 is equal to the measure of angle 𝐸𝐶𝐴?
In this question, we’re given a
construction and we need to determine which of five given options is correct in the
construction. We need to start by sketching
triangle 𝐴𝐵𝐶 by using the given side lengths of the triangle. We could do this by noting that the
lengths of three, four, and five centimeters make a Pythagorean triple. This is enough to know that it is a
right triangle with a hypotenuse of five centimeters by using the side-side-side
congruency criterion.
However, it is not necessary to
notice this to construct this triangle. In general, we can construct a
triangle from its lengths by using circles. We start by drawing one of its
sides. Let’s say 𝐴𝐵, which has a length
of three centimeters. We can then trace a circle of
radius four centimeters centered at 𝐵 and a circle of radius five centimeters
centered at point 𝐴. Either point of intersection
between these circles can be our point 𝐶. We choose 𝐶 as shown. We note that triangle 𝐴𝐵𝐶 has
the desired lengths.
We now want to extend the line
segment 𝐵𝐶 to be the ray from 𝐵 through 𝐶 so that we can find the point 𝐷 on
this ray to make the angle 𝐷𝐶𝐸 congruent to the angle 𝐶𝐵𝐴. To do this, let’s start by
highlighting the angle 𝐶𝐵𝐴 that we want to be congruent to the angle 𝐷𝐶𝐸. We know that this is a right angle
since this is a Pythagorean triple. However, it is not necessary to use
this to answer the question.
Instead, we can duplicate this
angle of vertex 𝐶 by using our construction for duplicating an angle at a
point. To duplicate this angle, we start
by tracing a circle at 𝐵 with radius less than three centimeters. We call the points of intersection
with the sides of the angle 𝐴 prime and 𝐶 prime. It is worth noting that it is
easier to choose a radius below 1.5 centimeters.
Next, we trace a circle of the same
radius centered at 𝐶. We label the point of intersection
between the ray from 𝐵 through 𝐶 and the circle that is above 𝐶 as point 𝐷. Now, we can duplicate the angle
shown at 𝐶 by tracing a circle of radius 𝐴 prime 𝐶 prime centered at 𝐷 and
labeling the point of intersection between the circles 𝐸 as shown. We then know that angle 𝐷𝐶𝐸 is
congruent to angle 𝐶𝐵𝐴. We can also see that line 𝐵𝐶 is a
transversal of the lines 𝐶𝐸 and 𝐴𝐵. And this transversal makes the same
angle with both lines. We can recall that this means that
the two lines must be parallel.
We can then note that the line
between 𝐴 and 𝐶 is also a transversal between these parallel lines. So the alternating angles it makes
with each line must be congruent. This gives us that angle 𝐸𝐶𝐴
must have the same measure as angle 𝐴, which we can see is option (A).
