Prove the Tangent-Chord Theorem using the provided steps.

Question

ScanSolve AI Tutor · Accepted Answer

To prove the Tangent-Chord Theorem, we will show that $\angle CBE = \angle BDE$ .

Construction.
Draw a diameter BF through the point of contact B. Connect F to E, forming chord EF.
Use properties of tangents and angles in a semicircle.
The tangent CAB is perpendicular to the radius at the point of contact B. Since BF is a diameter, it contains the radius.
$\angle CBF = 90^\circ$
The angle subtended by a diameter at any point on the circumference is $90^\circ$ . Therefore, $\angle BEF$ is an angle in a semicircle.
$\angle BEF = 90^\circ$
In $\triangle BEF$ , the sum of angles is $180^\circ$ . Since $\angle BEF = 90^\circ$ , the other two angles are complementary:
$\angle EBF + \angle BFE = 90^\circ$
From this, we can express $\angle BFE$ :
$\angle BFE = 90^\circ - \angle EBF$
Now, consider $\angle CBE$ . We know $\angle CBF = 90^\circ$ .
$\angle CBE = \angle CBF - \angle EBF$ $\angle CBE = 90^\circ - \angle EBF$
Comparing the expressions for $\angle CBE$ and $\angle BFE$ , we see they are equal:
$\angle CBE = \angle BFE$
Use angles subtended by the same arc.
Angles subtended by the same arc at the circumference are equal. Both $\angle BFE$ and $\angle BDE$ are subtended by the arc BE.
$\angle BFE = \angle BDE$
Conclude the proof.
From Step 2, we established that $\angle CBE = \angle BFE$ .
From Step 3, we established that $\angle BFE = \angle BDE$ .
Therefore, by transitivity, we can conclude:
$\boxed{\angle CBE = \angle BDE}$
This proves the Tangent-Chord Theorem.

Answer