Let A, C be the endpoints of the diameter of a circle and B an arbitrary point on the circle. Using the

slopes of secant lines show that \ABC is a right angle. You can assume the circle is centered at the
origin.

we can without loss of generality assume the radius is 1, and the diameter is on the x-axis. the coordinates of B are (cosθ,sinθ).

slope of AB = sinθ/(1+cosθ)
slope of CB = -sinθ/(1-cosθ)

product of slopes: -sin^2 θ/(1-cos^2 θ) = -1

so, the lines are perpendicular.

Steve your a boss for asking this question, you just saved me 2 hours of my life trying to figure it out, YEEEE

Kole*

To show that angle ABC is a right angle, we can use the concept of slopes of secant lines.

First, let's consider the coordinates of the given points. Let A = (xA, yA), C = (xC, yC), and B = (xB, yB).

Since the circle is centered at the origin, the coordinates of A and C are on the circle, which means they satisfy the equation of the circle: x^2 + y^2 = r^2.

Since A and C are the endpoints of the diameter, their coordinates can be represented as (r, 0) and (-r, 0) respectively. Therefore, we have:

A = (r, 0) and C = (-r, 0).

Now, let's consider the slope of the secant lines AB and BC.

The slope of a line passing through two points (x1, y1) and (x2, y2) can be calculated using the formula:

slope = (y2 - y1) / (x2 - x1).

For line AB, the coordinates are A = (r, 0) and B = (xB, yB). Therefore, the slope of AB is:

slope_AB = (yB - 0) / (xB - r).

For line BC, the coordinates are B = (xB, yB) and C = (-r, 0). Therefore, the slope of BC is:

slope_BC = (0 - yB) / (-r - xB).

To prove that angle ABC is a right angle, we can show that the product of the slopes of AB and BC is -1.

slope_AB * slope_BC = [(yB - 0) / (xB - r)] * [(0 - yB) / (-r - xB)].

We can observe that in both fractions, the numerator of the first fraction cancels out with the denominator of the second fraction and vice versa.

This simplifies the expression to:

slope_AB * slope_BC = [(-yB) / (xB - r)] * [(yB) / (r + xB)].

Now, notice that the negative sign in front of yB cancels out because the product of two negative numbers is positive.

slope_AB * slope_BC = [(yB) / (xB - r)] * [(yB) / (r + xB)].

This expression can be rewritten as:

slope_AB * slope_BC = [(yB)^2 / ((xB - r)(r + xB))].

Since we assumed that (xB, yB) lies on the circle, it satisfies the equation x^2 + y^2 = r^2. Therefore, we can replace (xB - r)(r + xB) with r^2 - xB^2.

slope_AB * slope_BC = [(yB)^2 / (r^2 - xB^2)].

Now, recall that (yB)^2 is always positive. Therefore, for the expression slope_AB * slope_BC to be -1 (indicating a right angle), the denominator (r^2 - xB^2) must be negative.

The terms r^2 and xB^2 are always positive, so for their difference (r^2 - xB^2) to be negative, xB^2 must be greater than r^2. However, this is not possible because (xB, yB) lies on the circle, which means it must satisfy x^2 + y^2 = r^2.

Since xB^2 cannot be greater than r^2, the denominator (r^2 - xB^2) cannot be negative. Therefore, the expression slope_AB * slope_BC cannot be -1, which implies that angle ABC is not a right angle.

Hence, the statement "angle ABC is a right angle" is incorrect.