Question

A triangle ABC is right angled at A. L is a point on BC such that AL $\perp$ BC. Prove that $\angle$ BAL =$\angle$ ACB.

Answer 1

Since $\Delta$ ABC is right-angled,

$A{B}^2+A{C}^2=B{C}^2$

or, $A{B}^2+A{C}^2=(BL+CL{)}^2$

or, $A{B}^2+A{C}^2=B{L}^2+C{L}^2+2.BL.CL$.......(1).

Again from $\Delta$ ALC and $\Delta$ ALB we have

$A{C}^2=A{L}^2+C{L}^2$......(2) and $A{B}^2=A{L}^2+B{L}^2$.....(3).

Adding (2) and (3) we get,

$A{C}^2+A{B}^2=2.A{L}^2+C{L}^2+B{L}^2$

or, $B{L}^2+C{L}^2+2.BL.CL=2.A{L}^2+B{L}^2+C{L}^2$

or, $A{L}^2=BL.CL$.......(4).

Now $\tan \angle ACB=\frac{AL}{CL}$.......(5) and

$\tan \angle ABL=\frac{AL}{BL}$

$\Rightarrow \tan ({90}^o-\angle BAL)=\frac{AL}{BL}$

$\Rightarrow \cot \angle BAL=\frac{AL}{BL}$......(6).

From (4) we have $\frac{AL}{CL}=\frac{BL}{AL}$

or, $\tan \angle ACB=\tan \angle BAL$[Using (5) and (6)]

or, $\angle ACB=\angle BAL$.