In the given figure,$PN$ and $QM$ are both perpendicular to line segment $PQ$. Also,$X$ is the midpoint of $PQ$ as well as $MN$. Prove that $\triangle PNX \cong \triangle QMX$.
(N/A) Given:
$1$. $PN \perp PQ$ and $QM \perp PQ$.
$2$. $X$ is the midpoint of $PQ$,so $PX = QX$.
$3$. $X$ is the midpoint of $MN$,so $NX = MX$.
To prove: $\triangle PNX \cong \triangle QMX$.
Proof:
In $\triangle PNX$ and $\triangle QMX$:
$1$. $PX = QX$ (Given,$X$ is the midpoint of $PQ$).
$2$. $NX = MX$ (Given,$X$ is the midpoint of $MN$).
$3$. $\angle PNX = \angle QMX$ (Since $PN \parallel QM$ because both are perpendicular to $PQ$,and $MN$ is a transversal,these are alternate interior angles).
Alternatively,using $SAS$ congruence criterion:
$1$. $PX = QX$ (Given).
$2$. $\angle P = \angle Q = 90^\circ$ (Given).
$3$. $NX = MX$ (Given).
Therefore,by $RHS$ congruence criterion (or $SAS$),$\triangle PNX \cong \triangle QMX$.
$1$. $PN \perp PQ$ and $QM \perp PQ$.
$2$. $X$ is the midpoint of $PQ$,so $PX = QX$.
$3$. $X$ is the midpoint of $MN$,so $NX = MX$.
To prove: $\triangle PNX \cong \triangle QMX$.
Proof:
In $\triangle PNX$ and $\triangle QMX$:
$1$. $PX = QX$ (Given,$X$ is the midpoint of $PQ$).
$2$. $NX = MX$ (Given,$X$ is the midpoint of $MN$).
$3$. $\angle PNX = \angle QMX$ (Since $PN \parallel QM$ because both are perpendicular to $PQ$,and $MN$ is a transversal,these are alternate interior angles).
Alternatively,using $SAS$ congruence criterion:
$1$. $PX = QX$ (Given).
$2$. $\angle P = \angle Q = 90^\circ$ (Given).
$3$. $NX = MX$ (Given).
Therefore,by $RHS$ congruence criterion (or $SAS$),$\triangle PNX \cong \triangle QMX$.
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$ABC$ is an isosceles triangle with $AB = AC$ and $D$ is a point on $BC$ such that $AD \perp BC$. To prove that $\angle BAD = \angle CAD$,a student proceeded as follows:
In $\triangle ABD$ and $\triangle ACD$:
$AB = AC$ (Given)
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