Question

In a square $ABCD$, its diagonals $AC$ and $BD$ intersect each other at point $O$. The bisector of angle $DAO$ meets $BD$ at point $M$ and the bisector of angle $ABD$ meet $AC$ at $N$ and $AM$ at $L$. Show that:
$\angle ONL+\angle OML={180}^o$
$\angle BAM=\angle BMA$
$ALOBisacyclicquadrilateral$.

Answer 1

$ABCD$ is a square whose diagonals $AC$ and $BD$ intersect each other at right angles at $O$.

$\left(i\right)\therefore \angle AOB=\angle AOD={90}^{\circ }$

In $△ANB$

$\angle ANB={180}^{\circ }-(\angle NAB+\angle NBA)$

$\Rightarrow \angle ANB={180}^{\circ }-({45}^{\circ }+\frac{{45}^{\circ }}{2})$ since $NB$ is bisector of $\angle ABD$

$\Rightarrow \angle ANB={180}^{\circ }-{45}^{\circ }-\frac{{45}^{\circ }}{2}$

$\Rightarrow \angle ANB={135}^{\circ }-\frac{{45}^{\circ }}{2}$

But $\angle LNO=\angle ANB$(vertically opposite angles)

$\therefore \angle LNO={135}^{\circ }-\frac{{45}^{\circ }}{2}$ ........$\left(1\right)$

Now in $△AMO,$

$\angle AMO={180}^{\circ }-(\angle AOM+\angle OAM)$

$\Rightarrow \angle AMO={180}^{\circ }-({90}^{\circ }+\frac{{45}^{\circ }}{2})$ since $MA$ is bisector of $\angle DAO$

$\Rightarrow \angle AMO={180}^{\circ }-{90}^{\circ }-\frac{{45}^{\circ }}{2}$

$\Rightarrow \angle AMO={90}^{\circ }-\frac{{45}^{\circ }}{2}$ ........$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ we get

$\angle LNO+\angle AMO={135}^{\circ }-\frac{{45}^{\circ }}{2}+{90}^{\circ }-\frac{{45}^{\circ }}{2}$

$\Rightarrow \angle LNO+\angle AMO={225}^{\circ }-{45}^{\circ }={180}^{\circ }$

$\Rightarrow \angle LNO+\angle AMO={180}^{\circ }$

$\left(ii\right)\angle BAM=\angle BAO+\angle OAM$

$\Rightarrow \angle BAM={45}^{\circ }+\frac{{45}^{\circ }}{2}=67{\frac{1}{2}}^{\circ }$

And $\Rightarrow \angle BMA={180}^{\circ }-(\angle AOM+\angle OAM)$

$\Rightarrow \angle BMA={180}^{\circ }-({90}^{\circ }+\frac{{45}^{\circ }}{2})$

$\Rightarrow \angle BMA={90}^{\circ }-\frac{{45}^{\circ }}{2}=67{\frac{1}{2}}^{\circ }$

$\therefore \angle BMA=\angle BAM$

$\left(iii\right)$ In quadrilateral $ALOB$

$\because \angle ABO+\angle ALO={45}^{\circ }+{90}^{\circ }+{45}^{\circ }={180}^{\circ }$

$\therefore ,ALOB$ is a cyclic quadrilateral.