Question

Question 9
A $\Delta ABC$ is right angled at A. L is a point on BC such that AL $\perp$ BC. Prove that $\angle BAL=\angle ACB$.

Answer 1

In $\Delta ABC,\angle A={90}^{\circ }andAL\perp BC$.
To prove: $\angle BAL=\angle ACB$
Proof:
In $\Delta ABCand\Delta \angle LAC,\angle BAC=\angle ALC\left[each{90}^{\circ }\right]\cdots \left(i\right)$
$and\angle ABC=\angle ABL\left[commonangle\right]\cdots \left(ii\right)$

On adding equations (i) and (ii), we get,
$\angle BAC+\angle ABC=\angle ALC+\angle ABL\cdots \left(iii\right)$
Again, in $\Delta ABC$,
$\angle BAC+\angle ACB+\angle ABC={180}^{\circ }$
[sum of all angles of a triangle is ${180}^{\circ }$]
$\Rightarrow \angle BAC+\angle ABC=180-\angle ACB$ . . . .(iv)
$in\Delta ABL,$
$\angle ABL+\angle ALB+\angle BAL={180}^{\circ }$
[sum of all angles of a triangle is ${180}^{\circ }$]
$\Rightarrow \angle ABL+\angle ALC={180}^{\circ }-\angle BAL[\because \angle ALC=\angle ALB={90}^{\circ }]...\left(v\right)$
On substituting the value from equations (iv) and (v) in Eq. (iii), we get,
${180}^{\circ }-\angle ACB={180}^{\circ }-\angle BAL$
$\angle ACB=\angle BAL$