Line segments $AB$ and $CD$ bisect each other at $P$. If $PA = PD$ and $PB = PC$,prove that $AC = BD$.
(N/A) Given: $AB$ and $CD$ bisect each other at $P$. This implies $PA = PB$ and $PC = PD$.
However,the problem states $PA = PD$ and $PB = PC$.
Consider $\triangle APC$ and $\triangle BPD$:
$1$. $PA = PD$ (Given)
$2$. $\angle APC = \angle BPD$ (Vertically opposite angles)
$3$. $PC = PB$ (Given)
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle APC \cong \triangle BPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AC = BD$.
However,the problem states $PA = PD$ and $PB = PC$.
Consider $\triangle APC$ and $\triangle BPD$:
$1$. $PA = PD$ (Given)
$2$. $\angle APC = \angle BPD$ (Vertically opposite angles)
$3$. $PC = PB$ (Given)
By $SAS$ (Side-Angle-Side) congruence criterion,$\triangle APC \cong \triangle BPD$.
Since the triangles are congruent,their corresponding parts are equal $(CPCT)$.
Therefore,$AC = BD$.
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