Hushar Mulga
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In the adjoining figure circles with centres X and Y touch each other at point Z

Chapter 3 – Circle – Text Book Solution

Problem Set 3 | Q 7 | Page 85

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`]

In the adjoining figure circles with centres X and Y touch each other at point Z.
solution

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]

Explanation:- 

The given construction is to draw segments XZ and YZ. The proof is as follows:

By the theorem of touching circles, since the circles with centers A and B touch at point Z, and X and Y are the points of contact of the circles with the line of centers, therefore points X, Z, and Y are collinear.

Thus, we have angle XZA congruent to angle BZY (opposite angles).

Let angle XZA = angle BZY = a.

Now, XA is congruent to XZ and YB is congruent to YZ, since they are radii of the same circles.

Therefore, angle XAZ = angle XZA = a (isosceles triangle theorem).

Similarly, angle BZY = angle ZBY = a (isosceles triangle theorem).

Therefore, from (I), (II), (III), we have angle XAZ = angle ZBY.

Therefore, radius XA is parallel to radius YB (alternate angle test).

Chapter 3 – Circle – Text Book Solution

Problem Set 3 | Q 7 | Page 85

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