Question

In the adjoining figure circles with centres X and Y touch each other at point Z. A secant passing through Z intersects the circles at points A and B respectively. Prove that, radius XA || radius YB. Fill in the blanks and complete the proof.

Construction: Draw segments XZ and YZ.

Proof:

By theorem of touching circles, points X, Z, Y are `square`.
∴ ∠XZA ≅ `square`                     ...(opposite angles)
Let ∠XZA = ∠BZY = a       ...(I)
Now, seg XA ≅ seg XZ      ...[Radii of the same circle]
∴∠XAZ = `square` = a                ...[isosceles triangle theorem](II)
Similarly, 

seg YB ≅ seg YZ            ...[Radii of the same circle]
∴∠BZY = `square` = a   ...[isosceles triangle theorem](III)
∴ from (I), (II), (III), 
∠XAZ = `square`
∴ radius XA || radius YZ     ...[`square`]

Answer 1

Construction: Draw segments XZ and YZ.
Proof:

By theorem of touching circles, points X, Z, Y are collinear.
∴ ∠XZA ≅ ∠BZY               ...(opposite angles)
Let ∠XZA = ∠BZY = a       .....(I)
Now, seg XA ≅ seg XZ      ...[Radii of the same circle]
∴∠XAZ = ∠XZA = a         ....[isosceles triangle theorem](II)
Similarly, seg YB ≅ seg YZ ....[Radii of the same circle]
∴∠BZY = ∠ZBY = a           ...[isosceles triangle theorem](III)
∴ from (I), (II), (III), 
∠XAZ = ∠ZBY
∴ radius XA || radius YB    ...[Alternate angle test]