`A B C` is a triangle in which `A B=A C\ a n d\ P` is a point on `A C` . Through `C` a line is drawn to intersect `B P` produced at `Q` such that `/_A B Q=\ /_A C Qdot` Prove that: `/_A Q C=90^0+1/2/_B A C`

Explanation

Answer
Given: `ABC` is a triangle in which `AB = AC` and `p` is point on `AC`.

Through `c` a line is drawn to intersect `BP` Produced at `Q `Such that, `/_ABQ=/_ACQ`.
`/_ABQ =/_ ACQ`
`=> A, B, C` and `Q`are concyclic points.
Theorem used---------------[If a line segment joining two points subtendes equal angles at two other points lying on the same side of the line segment, then the four are concyclic]
NOW,

`AB = AC`
`=>/_ACB=/_ABC `
But in triangle `ABC`

`/_BAC+/_ ACB + /_ABC = 180^@`

`=> /_BAC + 2/_ACB = 180^@` (`:. /_ACB =/_ ABC`)
`=> /_ACB = 90^@ - 1/2 /_BAC`..... `(1)`
Also,
` /_ACB = /_AQB and /_BAC = /_BQC` (angles in the same segment are equal)
`=>/_BAC + /_ACB =/_ AQB + /_BQC`

`=>/_BAC + /_ACB = /_AQC`

`/_ACB=/_ AGC- /_BAC` ...`(2)`
from equations `(1)` and `(2)`
we have,
`=>/_AQC-/_BAC = 90 ^@-1/2 /_BAC`
`=> /_AQC = 90^@ + 1/2 /_BAC`.
Hence proved.