Video Transcript
In the given figure, if 𝐴𝐷 is not
equal to 𝐴𝐵 which is not equal to 𝐵𝐶, which of the following is a tangent to the
circle that passes through the vertices of the triangle 𝐴𝐵𝐸? Is it (A) segment 𝐴𝐷, or (B)
segment 𝐵𝐶, (C) ray 𝐵𝑌, (D) segment 𝐸𝐶, or (E) segment 𝐸𝐷?
To understand the question, let’s
first annotate our diagram to show where the circle that passes through the vertices
of the triangle 𝐴𝐵𝐸 sits. Here we can see both the triangle
𝐴𝐵𝐸 and the circumcircle that passes through all three of these vertices.
Now, considering our options for
potential tangents to this new circle, it seems plausible that segment 𝐴𝐷 could be
a tangent; that’s option (A). It also seems plausible that
segment 𝐵𝐶 could be a tangent, that’s option (B). However, we can eliminate the ray
𝐵𝑌, that’s option (C), as it passes directly through the circle. We can eliminate line segment 𝐸𝐶
as this is an extension of the chord 𝐴𝐸, and with the same reasoning, we can
eliminate the line segment 𝐸𝐷.
To narrow this down further between
our remaining two choices, options (A) and (B), we’ll need to consider what we can
say about the angles inside this figure. We can see in our figure that
segment 𝐵𝐷 is parallel to the ray 𝑋𝐶 and ray 𝑋𝐶 is tangent to the larger
circle. Since segment 𝐵𝐶 intersects two
parallel lines, it’s a transversal, and so the measures of angles 𝑋𝐶𝐵 and 𝐶𝐵𝐸
will be equal. From there, since triangle 𝐴𝐶𝐵
is inscribed in the larger circle and point 𝐶 is a point of tangency, by the
alternate segment theorem, angle 𝐵𝐴𝐶 will be equal in measure to angle
𝑋𝐶𝐵.
Remember, the alternate segment
theorem tells us that if 𝐴 and 𝐵 are two points on a circle and 𝐶 is a tangent
point on the circle, then the angle of tangency 𝐵𝐶𝑆 is equal to the angle 𝐵𝐴𝐶
in the alternate segment. And because the measure of angle
𝐶𝐵𝐸 has been shown to be equal to the measure of angle 𝐵𝐴𝐸, by the converse of
the alternate segment theorem, the segment 𝐵𝐶 must be tangent to the circle that
passes through the vertices of the triangle 𝐴𝐵𝐸. So by using properties of parallel
lines and the alternate segment theorem and its converse, we’ve been able to show
that line segment 𝐵𝐶 is a tangent to the circle that passes through the vertices
of triangle 𝐴𝐵𝐸.
Now for completion, thinking about
the chord 𝐴𝐷, we know that it’s not equal in length to the chord 𝐵𝐶. And if we consider the fact that
two angles subtended by the same arc are equal, our angles 𝐴𝐵𝐷 and 𝐴𝐶𝐷 have
equal measure and similarly for angles 𝐵𝐴𝐶 and 𝐵𝐷𝐶. But the two pairs are not of equal
measure, since the lengths of the arcs from which they subtend are not equal. We see in our larger circle, this
means that the measure of angle 𝐴𝐵𝐷, let’s call this 𝜑, would not be equal to
that of angle 𝐵𝐴𝐶, which is our angle 𝜃.
So now, making some space, let’s
suppose that line segment 𝐴𝐷 is a tangent to the circle passing through the
vertices 𝐴𝐵 and 𝐸. Then, by the alternate segment
theorem, the angles 𝐴𝐵𝐷, that’s 𝜑, and 𝐷𝐴𝐸 should have equal measure. Now using again the fact that
angles subtended by the same arc are equal, we have that the measure of angle 𝐷𝐴𝐶
equals that of 𝐶𝐵𝐷. But we’ve already shown that the
measure of angle 𝐶𝐵𝐸, and therefore 𝐶𝐵𝐷, is equal to 𝜃. And this must mean that angle
𝐷𝐴𝐶, that’s 𝐷𝐴𝐸, is also equal to 𝜃. But angle 𝐷𝐴𝐸 is 𝜑, which we’ve
already seen does not equal 𝜃. And so we have a contradiction. Hence, we’ve shown by contradiction
that the line segment 𝐴𝐷 cannot be a tangent to the circle passing through the
vertices of the triangle 𝐴𝐵𝐸. And this discounts option (A).
Therefore, only option (B), that’s
the line segment 𝐵𝐶, is a tangent to the circle passing through the vertices 𝐴,
𝐵, and 𝐸.
