Question

In the given figure, 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.

Answer 1

Circles with centres X and Y touch internally at point Z.
Join YZ.

By theorem of touching circles, points Y, X, Z are collinear.
Now, seg XA ≅ seg XZ            (Radii of circle with centre X)
∴∠XAZ = ∠XZA                   (Isosceles triangle theorem)              .....(1)
Similarly, seg YB ≅ seg YZ        (Radii of circle with centre Y)
∴∠BZY = ∠ZBY                       (Isosceles triangle theorem)          .....(2)
From (1) and (2), we have
∠XAZ = ∠ZBY
If a pair of corresponding angles formed by a transversal on two lines is congruent, then the two lines are parallel.
∴ seg AX || seg BY                     (Corresponding angle test)