In the diagram, $AT$ is the tangent to the circle at $T$. $ABC$ is a straight line and \(AB=TB

Question

In the diagram, $AT$ is the tangent to the circle at $T$. $ABC$ is a straight line and $AB=TB$. Prove that $\angle BAT=\angle BCT$, $\Delta ACT$ is similar to $\Delta ATB$, $AT^2=BT \times AC$.

Accepted Answer

(a) Proved: $\angle BAT=\angle BCT$ (b) Proved: $\triangle ACT\sim\triangle ATB$ (c) Proved: $AT^2=BT\times AC$

Proofs in Plane Geometry - Alternate Segment Theorem

Proofs in Plane Geometry - Alternate Segment Theorem

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