Question

In the adjoining figure, circles with centres X and Y touch internally at point Z. Seg BZ is a chord of bigger circle and it intersects smaller circle at point A. Prove that, seg AX || seg BY.

Answer 1

Given: X and Y are the centres of the circle. 

To prove: seg AX || seg BY 

Proof: 

In ∆XAZ, 

seg XA ≅ seg XZ [Radii of the same circle] 

∴ ∠XZA ≅ ∠XAZ ………… (i) [Isosceles triangle theorem] Also, in ∆YBZ, 

seg YB ≅ seg YZ [Radii of the sanie circle] 

∴ ∠YZB ≅∠YBZ [Isosceles triangle theorem]

∴ ∠XZA ≅ ∠YBZ ………….. (ii) [Y – X – Z,B – A – Z] 

∴ ∠XAZ ≅ ∠YBZ [From (i) and (ii)] 

∴ seg AX || seg BY [Corresponding angles test]