S is the centre of the circumcircle of `Delta ABC` and if `AD bot BC`, prove that ` angle BAD= angle SAC`.

Explanation

S is the centre of the circum-circle of `Delta ABC`.
`:. SA=SB ne SC`.
Also, `AD bot BC, angle ADB =90^(@), :. angle ABD+ angle BAD =90^(@)`
or, `angle ABD-90^(@)-angle BAD........(1)`
Now, in the circle with centre S, the central angle produced by the chord AD `= angle ASC` and angle in circle `angleABC`
`:. ASC=2 angle ABC .......(2)`
But `SA=SC[ :.` radii of same circle]
`:. angle SAC= angle SCA.......(3)`
Then, `angle ASC+ angle SAC+ angleSCA= 180^(@)`
or, `angle ASC+ angle SAC+ angleSAC= 180^(@)[ "from"(3)]`
or `angle ASC+2 angle SAC=180^(@)`
or, `2 angle ABC+ 2 angle SAC =180^(@)` [ by (2)]
or, `angle ABC+ angle SAC =90^(@)` [ dividing by 2]
or, `angle ABD+ angle SAC =90^(@) [ :. angle ABC and angle ABD` are same angle]
`or, 90^(@)- angle angle BAD+ angle SAC=90^(@)` [ from (1)]
or, `angle SAC- angle BAD=0`, or ` angle SAC= angle BAD`
Hence `angle ABD= angle SAC`