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1.) Prove that triangles ABC and EDC are congruent.
Given:
Angle ABC is a right angle.
Angle EDC is a right angle.
C is the midpoint of BD.
AB=ED​

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Lorxis
I wrote my STATEMENTS and REASONS as follows:
STATEMENT: 1. Line segments AE & DB intersect at C.
REASON: 1. Given.
STATEMENT: 2. Line segments AC & EC are congruent.
REASON: 2. Given.
STATEMENT: 3. Line segments BC & DC are congruent.
REASON: 3. Given.
STATEMENT: 4. Angles ACB & ECD are congruent.
REASON: 4. Vertical angles are congruent (Theorem) VERTICALLY OPPOSITE ANGLES ARE EQUAL..THAT IS THE CORRECT STATEMENT.
STATEMENT: 5. Triangles ABC & EDC are congruent.
REASON: 5. (SSS are congruent to SSS). If three sides of one triangle are congruent, respectively, to three sides of a second triangle, then the triangles are congruent. (Postualte) ...NO THEOREM IS SAS..2 SIDES AND INCLUDED REPEAT INCLUDED ANGLES ARE EQUAL RESPECTIVELY THE CORRESPONDING SIDES AND INCLUDED ANGLE , THEN THE 2 TRIANGLES ARE CONGRUENT.
I'm not sure if Step 4 is correct or if I can even write that statement.IT IS OK BUT FOR THE SLIGHT MODIFICATION SUGGESTED.
REASON 5 IS TO BE CHANGED AS GIVEN .

Following are the proving to [tex]\Delta ABC\ and \ \Delta EDC[/tex] that are congruent:

Given:

Please find the complete question.

To find:

Prove

Solution:

Solving the choices in the given points:

When the [tex]\bold{\angle}[/tex]ABC is a right angle then:

[tex]\bold{m \angle ABC = 90^{\circ}}[/tex]

When the [tex]\bold{\angle}[/tex] EDC is a right angle then:

[tex]\bold{m \angle EDC = 90^{\circ}}[/tex]

[tex]\bold{\angle ABC \cong \angle EDC}[/tex]

When C is the midpoint of  [tex]\overline{BD}.[/tex]

[tex]\overline{BC} \cong \overline{DC}\\\\\overline{AB}\cong \overline{ED}[/tex]

So,  

[tex]\bold{\Delta ABC \cong \Delta EDC \ \ \ \ \ \ (S A S)}[/tex]

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