### Video Transcript

Draw triangle π΄π΅πΆ where π΄π΅
equals three centimeters, π΅πΆ equals four centimeters, and π΄πΆ equals five
centimeters. Point π· lies on the ray from π΅
through πΆ such that π· is not on line segment π΅πΆ. Draw angle π·πΆπΈ congruent to
angle πΆπ΅π΄, where πΈ is on the upper side of line segment π΅πΆ. Which of the following is true? Option (A) the measure of angle π΄
equals the measure of angle πΈπΆπ΄. Option (B) the measure of angle π΄
is equal to the measure of angle π΅πΆπ΄. Option (C), the measure of angle π΄
equals the measure of angle πΈπΆπ·. Or is it option (D) the measure of
angle π΅ is equal to the measure of angle πΈπΆπ΄?

In this question, weβre 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 π΄π΅πΆ 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. Letβs say π΄π΅, which has a length
of three centimeters. We can then trace a circle of
radius four centimeters centered at π΅ and a circle of radius five centimeters
centered at point π΄. Either point of intersection
between these circles can be our point πΆ. We choose πΆ as shown. We note that triangle π΄π΅πΆ has
the desired lengths.

We now want to extend the line
segment π΅πΆ to be the ray from π΅ through πΆ so that we can find the point π· on
this ray to make the angle π·πΆπΈ congruent to the angle πΆπ΅π΄. To do this, letβs start by
highlighting the angle πΆπ΅π΄ that we want to be congruent to the angle π·πΆπΈ. 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 πΆ by using our construction for duplicating an angle at a
point. To duplicate this angle, we start
by tracing a circle at π΅ with radius less than three centimeters. We call the points of intersection
with the sides of the angle π΄ prime and πΆ 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 πΆ. We label the point of intersection
between the ray from π΅ through πΆ and the circle that is above πΆ as point π·. Now, we can duplicate the angle
shown at πΆ by tracing a circle of radius π΄ prime πΆ prime centered at π· and
labeling the point of intersection between the circles πΈ as shown. We then know that angle π·πΆπΈ is
congruent to angle πΆπ΅π΄. We can also see that line π΅πΆ is a
transversal of the lines πΆπΈ and π΄π΅. 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 π΄ and πΆ 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 πΈπΆπ΄
must have the same measure as angle π΄, which we can see is option (A).