# Thread: proving ab subtends c at right angle

1. ## proving ab subtends c at right angle

A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.

I have tried drawing it so i can understand the problem better in order to answer the question. but looking at the sketch. It creates an isosceles
triangle not a right angle

I have tried a different sketch;

the problem with this sketch is that c is not the positive y-axis.

any tips so i can figure it out
thanks

2. ## Re: proving ab subtends c at right angle

A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.
What kind of triangles are BOC and AOC? How do you know?

What is the measure of angle BCO? ACO? ... so, what is the measure of angle BCA?

Angle inscribed in a semicircle - Math Open Reference

3. ## Re: proving ab subtends c at right angle

Originally Posted by bigmansouf
A circle, radius two units, center the origin , cuts the x-axis at A and B and cuts the positive y-axis at C. Prove that AB subtends a right angle at C.
You are being asked to prove analytically that any angle inscribed in a semi-circle is a right angle.
The x-intercepts are $A: (-2,0)~\&~B: (2,0)$. Any point of the upper semi-circle is $C: (a,\sqrt{4-a^2}),~-2\le a\le 2$

The slopes $S(\overline{AC})=\dfrac{\sqrt{4-a^2}}{a+2}~\&~S(\overline{BC})=\dfrac{\sqrt{4-a^2}}{a-2}$

Can you show that $S(\overline{AC})\cdot S(\overline{BC})=-1~?$

4. ## Re: proving ab subtends c at right angle

Originally Posted by bigmansouf
I have tried drawing it so i can understand the problem better in order to answer the question. but looking at the sketch. It creates an isosceles triangle not a right angle
Ever heard of a right-isosceles triangle (45-45-90)?