Question
If $AP$ bisects $\angle BAC$ and $M$ is any point on $AP,$ prove that the perpendiculars drawn from $M$ to $AB$ and $AC$ are equal.
✓
Answer
From $M$, draw $M L$ such that $M L$ is perpendicular to $A B$ and $M N$ is perpendicular to $\mathrm{AC}$
In $\triangle \mathrm{ALM}$ and $\triangle \mathrm{ANM}$
$\angle \mathrm{LAM}=\angle \mathrm{MAN} \ldots[\because \mathrm{AP}$ is the bisector of $\mathrm{BAC}]$
$ \angle \mathrm{ALM}=\angle \mathrm{ANM}=90^{\circ} \ldots[\because \mathrm{ML} \perp \mathrm{AB}, \mathrm{MN} \perp \mathrm{AC}]$
$ \mathrm{AM}=\mathrm{AM} \ldots[$ Common $]$
$\therefore$ By Angel$-$Angel$-$Side criterion of congruence,
$\triangle \mathrm{ALM} \cong \triangle \mathrm{ANM}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \mathrm{ML}=\mathrm{MN} \ldots[$ c. p.c.t $]$
Hence proved.
In $\triangle \mathrm{ALM}$ and $\triangle \mathrm{ANM}$
$\angle \mathrm{LAM}=\angle \mathrm{MAN} \ldots[\because \mathrm{AP}$ is the bisector of $\mathrm{BAC}]$
$ \angle \mathrm{ALM}=\angle \mathrm{ANM}=90^{\circ} \ldots[\because \mathrm{ML} \perp \mathrm{AB}, \mathrm{MN} \perp \mathrm{AC}]$
$ \mathrm{AM}=\mathrm{AM} \ldots[$ Common $]$
$\therefore$ By Angel$-$Angel$-$Side criterion of congruence,
$\triangle \mathrm{ALM} \cong \triangle \mathrm{ANM}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \mathrm{ML}=\mathrm{MN} \ldots[$ c. p.c.t $]$
Hence proved.
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