A triangle `A B C` is right angles at `AdotA L` is drawn perpendicular to `B C` . Prove that `/_B A L=\ /_A C B`

Explanation

In △ABL,
`∠BAL+∠ALB+∠B=180@`
`∠BAL+90@ +∠B=180@`
`∠BAL=90@−∠B` ......(1)
In `△ABC,
∠A+∠B+∠C=180@
∠B+∠C=90@
∠C=90@−∠B
∠ACB=90@−∠B `.......(2)'
From '1' and '2', we get,
`∠BAL=∠ACB`