Two circle with centres `A` and `B` intersect at `C` and `Ddot` Prove that `/_A C B=/_A D Bdot`

Explanation

Given: Two circles with centres `A` and `B` intersect at `C` and `D`. To Prove : `∠ACB=∠ADB`
Construction: Join `AC`, `AD`, `BC`, `BD` and `AB`.
Proof : In `∆ACB` and `∆ADB`,
`AC=AD` [Radii of the same circle]
`BC=BD` [Radii of the same circle]
`AB=AB` [Common]
`∴∆ACB≅∆ADB` [By `SSS`]
`∴∠ACB=∠ADB` [By `CPCT`]
Hence Proved.