Question
$A$ line segment $AB$ is bisected at point $P$ and through point $P$ another line segment $PQ$, which is perpendicular to $AB$, is drawn. Show that: $QA = QB.$
✓
Answer
Given: $A \triangle A B Q$ in which $A B$ is bisected at $P$
$\mathrm{PQ}$ is perpendicular to $\mathrm{AB}$
WE need to prove that
$Q A=Q B$
Proof:
In $\triangle \mathrm{APQ}$ and $\triangle \mathrm{BPQ}$
$\mathrm{AP}=\mathrm{PB} \ldots[\mathrm{P}$ is the mid$-$point of $\mathrm{AB}]$
$\angle \mathrm{APQ}=\angle \mathrm{BPQ}=90^{\circ} \ldots[\mathrm{PQ}$ is perpendicular to $\mathrm{AB}]$
$\mathrm{PQ}=\mathrm{PQ} \ldots[$ Common $]$
$\therefore$ By Side$-$Angel$-$Side criterion of congruence,
$\triangle \mathrm{APQ} \cong \triangle \mathrm{BPQ}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \mathrm{QA}=\mathrm{QB} \ldots[\text { c.p.c.t ] }$
$\mathrm{PQ}$ is perpendicular to $\mathrm{AB}$
WE need to prove that
$Q A=Q B$
Proof:
In $\triangle \mathrm{APQ}$ and $\triangle \mathrm{BPQ}$
$\mathrm{AP}=\mathrm{PB} \ldots[\mathrm{P}$ is the mid$-$point of $\mathrm{AB}]$
$\angle \mathrm{APQ}=\angle \mathrm{BPQ}=90^{\circ} \ldots[\mathrm{PQ}$ is perpendicular to $\mathrm{AB}]$
$\mathrm{PQ}=\mathrm{PQ} \ldots[$ Common $]$
$\therefore$ By Side$-$Angel$-$Side criterion of congruence,
$\triangle \mathrm{APQ} \cong \triangle \mathrm{BPQ}$
The corresponding parts of the congruent triangles are congruent.
$\therefore \mathrm{QA}=\mathrm{QB} \ldots[\text { c.p.c.t ] }$
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