In figure 3.88, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it itersects smaller circle at point A. Prove that, seg AX ∥ seg BY.
Figure: X is the centre of the smaller circle and Y is the centre of the bigger circle; Z–A–B.
Answer: \(\angle XAZ = \angle XZA = \angle YBZ\), which are corresponding angles, so AX ∥ BY.
Draw seg XZ and seg YZ. By the theorem of touching circles, Z, X, Y are collinear, with X on seg ZY.
In the smaller circle, XA = XZ (radii), so
\[\angle XAZ = \angle XZA\quad \text{(I)}\]In the bigger circle, YB = YZ (radii), so
\[\angle YBZ = \angle YZB\quad \text{(II)}\]A lies on seg ZB and X lies on seg ZY, so \(\angle XZA\) and \(\angle YZB\) are the same angle. From (I) and (II), \(\angle XAZ = \angle YBZ\).
These are corresponding angles made by transversal ZB with seg AX and seg BY. \(\therefore\) seg AX ∥ seg BY.
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