Given In `DeltaABC, /_A = 90^(@)` and AL `bot`BC
To prove `/_BAL = /_ACB`
Proof In `DeltaABC and DeltaLAC," " /_BAC = /_ALC " " [each 90^(@)] …..(i)`
and `" " /_ABC = /_ABL " " `[common angle] ....(ii)
On adding Eqs. (i) and (ii), we get `/_BAC +/_ABC + /_ALC + /_ABL " " .....(iii)`
Again, in `Delta ABC, /_BAC +/_ACB + /_ABC = 180^(@) " ".....(iv)`
[sum of all angles of a triangles is `180^(@)`]
`rArr "IN" DeltaABC, /_ABL +/_ALB + /_BAL 180^(@)" " `[sum of all angles of a triangle is `180^(@)`]
`rArr " "/_ABL +/_ALC = 180^(@)/_BAL " " [:./_ALC = /_ALB = 90^(@)]` ...(V)
On substituting the value from Eqs. (iv) and (v) in Eq. (iii), we get
`180^(@) -/_ ACB = 180^(@)-/_`BAL
`rArr" "/_ ACB =/_BAL`
Hence proved.
