I is the centre of the incircle of `Delta ABC,` produced AI intersects the circumercile of that triangle at the point P. Prove that `PB=PC=PI`

Answer

page 279

Explanation

I is the incentre of `Delta ABC`.
`:. angle IBC= angle IBC,angle IAB= angleIAC and angleICB = angleICA`
`:. angleIAB= angle IAC., :. anglePAB= anglePAC`
`:. B=PC` [ by the converse thorem of theore-35 ......(i)
Again, `angle BCP and angleBAP` are two angles in circle produced by the are `overset(frown)(PB)`.
`:.` by theorem, 35 `angleBCP= angleBAP`
or, `angle PBC= angle BAP [ :. PB =PC, :. angle BCP= PBC]`
Now, external `angle BIP of Delta AIB` = internally opposite
`(angle IAB+ angle IBA)= angle IAB+angleIBC`
`= angleBAP+ angleIBC [ :. angleIAB and angle BAP` are same angles]
`= angle PBC+ angle IBC]`
`[ :. anglePBC= angle BAP]=angle IBP`
`:. angle BIP+ angle IBP`
`:.` from (1) and (2) we get , `PB=PC=PI`