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 \[\boxed{\text{collinear}}\].

∴ ∠XZA ≅ \[\boxed{\text{∠BZY}}\]   ...(Opposite angles)

Let ∠XZA = ∠BZY = a   ...(i)

Now, seg XA ≅ seg XZ   ...[Radii of the same circle]

∴∠XAZ = \[\boxed{\text{∠XZA}}\] = a   ...[Isosceles triangle theorem] (ii)

Similarly, seg YB ≅ seg YZ   ...[Radii of the same circle]

∴∠BZY = \[\boxed{\text{∠ZBY}}\] = a   ...[Isosceles triangle theorem] (iii)

∴ from (i), (ii), (iii),

∠XAZ = \[\boxed{\text{∠ZBY}}\]

∴ radius XA || radius YB   ...\[\boxed{\text{[Alternate angle test]}}\]