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In the figure, DE`|""|` QR and BP are bisectors of `/_`EAB and `/_`RBA, respectively. Find `/_`APB. |
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Answer» 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^(@)` |
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