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Angle of triangle inside circle

Hi there,

In the above diagram:

Q is the center of the circle, PAT is a tangent to the circle, PR is parallel to AC, angle CAT = x. Prove that angle ABC = x.

I started off by going 90 - x to find CAB. Then used co-interior angles to find AQR, and then found out BQR.

I'm just curious, in this question, can I assume that QRB and ACB are right angled triangles? Becuase if they were, I would think the question would state that. How would I solve this?

John Omielan
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chris
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  • Triangle $QRB$ is not a right triangle. But the line $PR$ seems to be irrelevant to what you have to prove. – David K Oct 16 '19 at 10:44

1 Answers1

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Thale's theorem states $\angle ACB$ is a right angle since $AB$ is a diameter. As $\angle BAT$ is also a right angle due to $PAT$ being tangent to the circle at $A$ (as stated in Tangent lines to circles, "The radius of a circle is perpendicular to the tangent line through its endpoint on the circle's circumference."), you have $\angle CAB = 90^{\circ} - x$ (as you already stated in your question) and, thus, $\angle ABC = x$ due to the sum of the angles in a triangle being $180^{\circ}$.

John Omielan
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  • So the angle BRQ is also right angled, right? – chris Oct 16 '19 at 03:00
  • Sorry, I meant angle B, the intersection point close to R, and Q? – chris Oct 16 '19 at 03:01
  • And how does PAT being a tangent make BAT right angled? – chris Oct 16 '19 at 03:03
  • @chris If by "the intersection point close to R, and Q" you mean the point where $BC$ and $QR$ intersect (let's call it point $S$), then it is right angled also because since $PR$ is parallel to $AC$ means that $\angle BSQ = \angle BCA = 90^{\circ}$. – John Omielan Oct 16 '19 at 03:07
  • @chris I didn't realize you didn't know about the radius of a circle to the point where a tangent line touches the circle is perpendicular to the tangent line. I've added a link to Wikipedia's article about this and their statement of it. I hope this explains it to you. – John Omielan Oct 16 '19 at 03:12