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?
Best Answer
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 $180^{\circ}$.