Chapter 4.4, Problem 25PPS

To determine

To prove:

  $\Delta EAD\cong \Delta DCE$ by two-column.

Explanation of Solution

Given Information:

  $\Delta EAB\cong \Delta DCB$ .

  

Proof:

The two-column proof can be given as:

Proof	Reason
$\Delta EAB\cong \Delta DCB$	Given.
$\overline{EA}\cong \overline{DC}$	CPCTC
$\overline{ED}\cong \overline{DE}$	Reflexive property.
$\overline{AB}\cong \overline{CB}$	CPCTC
$\overline{DB}\cong \overline{EB}$	CPCTC
$\begin{array}{l}AB=CB\\ DB=EB\end{array}$	Definition of congruent segments.
$AB+DB=CB+EB$	Addition property.
$\begin{array}{l}AD=AB+DB\\ CE=CB+EB\end{array}$	Segment addition.
$AD=CE$	Substitution property.
$\overline{AD}\cong \overline{CE}$	Definition of congruent segments.
$\Delta EAD\cong \Delta DCE$	SSS − Side, Side, Side.