triangle ABC is right angled at A and AL _|_ BC . Prove that <BAL= <ACB

Solution :

We form our diagram from given information , As :

Now from angle sum property of triangle we get in $∆$ ABC

$\angle$ BAC  +  $\angle$ ACB + $\angle$ ABC  =  180$°$

90$°$ + $\angle$ ACB + $\angle$ ABC  =  180$°$                                    (  Given ABC right angled at A )

$\angle$ ACB + $\angle$ ABC  =  90$°$ 

$\angle$ ACB   =  90$°$ - $\angle$ ABC                                                     ------ ( 1 )

And $∆$ ALB

$\angle$ ALB  +  $\angle$ BAL + $\angle$ ABL  =  180$°$

90$°$ + $\angle$ BAL + $\angle$ ABC  =  180$°$                                    (  Given AL $\perp$ BC  and $\angle$ ABL = $\angle$ ABC (  Same angles ) )

$\angle$ BAL + $\angle$ ABC  =  90$°$ 

$\angle$ BAL   =  90$°$ - $\angle$ ABC                                                     ------ ( 2 )
Now, using  Euclid's axiom, things which are equal to same thing are equal to one another,
From equation (1)and (2) we get

$\angle$ BAL = $\angle$ ACB                                                                 ( Hence proved )