In the figure, DE`|""|` QR and BP are bisectors of `/_`EAB and `/_`RBA, respectively. Find `/_`APB.

Explanation

Given, DE`||` OR and AP and PB are the bisectors of `/_`EAB and `/_`RBA, respectively.
We know that. Interior angles on the same side of transversal are supplementary.
`:." " /_EAB + /_RBA = 180^(@)`
`rArr" " 1/2/_EAB +1/2/_RBA = 180^(@)/2" " ` [dividing both sides by 2]
`rArr" " 1/2/_EAB +1/2/_RBA = 90^(@)" "`...(i)
Since, AP and BP are the bisectors of `/_EAB and /_`RBA, respectively.
`:. " " /_BAP = 1/2 /_`EAB ....(ii)
and `" " /_ABP = 1/2 /_RBA " "` .........(iii)
On adding Eqs. (ii) and (iii), we get
`/_BAP + /_ABP = 1/2/_EAB +1/2/_`RBA
From Eqs. (i),
`rArr" "/_BAP + /_ABP = 90^(@) " "....(iv)`
In `DeltaAPB, " "/_BAP + /_ABP + /_APB = 180^(@) " "["sum of all angles of a triangle is" 180^(@)]`
`rArr" "90^(@)+/_APB = 180^(@) " "["from Eq".(iv)]`
`rArr" "/_APB = 180^(@)-90^(@)=90^(@)`