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प्रश्न
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.
उत्तर
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)
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∴ ∠XZA ≅ `square` ...(opposite angles)
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