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)