Prove that a diameter \(AB\) of a circle bisects every chord that is parallel to the tangent at \(A\).
Bisected.
Step 1: Draw a circle with center \(O\). Mark a point \(A\) on the circle.
Step 2: Draw the diameter \(AB\) of the circle, passing through \(A\) and the center \(O\).
Step 3: At point \(A\), draw a tangent to the circle. By definition, a tangent touches the circle at exactly one point.
Step 4: From geometry, the tangent at \(A\) is always perpendicular to the radius \(OA\).
Step 5: Now, draw a chord (say \(CD\)) of the circle such that it is parallel to the tangent at \(A\).
Step 6: Since \(CD\) is parallel to the tangent at \(A\), and the tangent is perpendicular to \(OA\), this means \(CD\) is also perpendicular to \(OA\).
Step 7: In circle geometry, a radius (or diameter) that is perpendicular to a chord always bisects the chord. That is, it divides the chord into two equal parts.
Step 8: Here, \(OA\) (or the diameter \(AB\)) is perpendicular to chord \(CD\). Therefore, diameter \(AB\) must pass through the midpoint of \(CD\).
Final Conclusion: Thus, the diameter \(AB\) bisects every chord of the circle that is parallel to the tangent at \(A\).