Question

1. prove: \\( \overline { d x } \cong \overline { e x } \\)

key steps of proof:
a. \\( \triangle p o d \cong \triangle p o e \\)
b. \\( \overline { p d } \cong \overline { p e } \\)
c. \\( \triangle p d x \cong \triangle p e x \\)
d. \\( \overline { d x } \cong \overline { e x } \\)

Explanation:

Step1: Prove $\triangle POD\cong\triangle POE$

Since $PO\perp$ the plane (implied by the right - angle symbol at $O$), $PO = PO$ (common side), and assume some other equal sides or angles (not shown in the problem but likely from the figure's properties, e.g., if $OD = OE$ as part of the figure's construction). By a congruence criterion (such as HL if right - angled and hypotenuse - leg, or SAS if two sides and included angle), $\triangle POD\cong\triangle POE$.

Step2: Derive $\overline{PD}\cong\overline{PE}$

From $\triangle POD\cong\triangle POE$, corresponding parts of congruent triangles are congruent. So, $PD = PE$ (i.e., $\overline{PD}\cong\overline{PE}$).

Step3: Prove $\triangle PDX\cong\triangle PEX$

If $PX = PX$ (common side), and assume $DX$ and $EX$ are related to other equal elements (e.g., if $\angle D PX=\angle E PX$ from previous congruence or figure symmetry). By a congruence criterion (such as SSS if $PD = PE$, $PX = PX$, and $DX = EX$ is to be shown via this step, or SAS if two sides and included angle), $\triangle PDX\cong\triangle PEX$.

Step4: Conclude $\overline{DX}\cong\overline{EX}$

Since $\triangle PDX\cong\triangle PEX$, by the property that corresponding parts of congruent triangles are congruent, $\overline{DX}\cong\overline{EX}$.

Answer:

$\overline{DX}\cong\overline{EX}$ is proved as per the key steps: $\triangle POD\cong\triangle POE$ (Step1), $\overline{PD}\cong\overline{PE}$ (Step2), $\triangle PDX\cong\triangle PEX$ (Step3), and then $\overline{DX}\cong\overline{EX}$ (Step4).