Question

the proof that $\triangle abc \cong \triangle cda$ is shown.
given: $\overline{ab} \cong \overline{cd}$ and $\overline{bc} \cong \overline{da}$
prove: $\triangle abc \cong \triangle cda$

what is the missing reason in the proof?

statements	reasons
2. quadrilateral $abcd$ is a $\parallelogram$	2. definition of parallelogram
3. $ab = cd$; $bc = da$	3. opposite sides of a parallelogram are $=$
4. $\overline{ac} = \overline{ac}$	4. reflexive property
5. $\triangle abc \cong \triangle cda$	5.?

options:

• perpendicular bisector theorem
• pythagorean theorem
• hl theorem
• sss congruence theorem

Explanation:

To determine the missing reason for proving \(\triangle ABC \cong \triangle CDA\), we analyze the given information:

• We know \(AB \cong CD\) and \(BC \cong DA\) (from the properties of the parallelogram and the given parallel sides).
• We also have \(AC \cong AC\) (the reflexive property, meaning a side is congruent to itself).

This gives us three pairs of congruent sides: \(AB \cong CD\), \(BC \cong DA\), and \(AC \cong AC\). The SSS (Side - Side - Side) congruence theorem states that if three sides of one triangle are congruent to three sides of another triangle, then the two triangles are congruent. The other options do not fit:

• The HL (Hypotenuse - Leg) theorem is for right - angled triangles, and there is no indication that these are right - angled triangles.
• The Pythagorean theorem relates to the sides of a right - angled triangle (\(a^{2}+b^{2}=c^{2}\)) and is not used for triangle congruence in this context.
• The perpendicular bisector theorem deals with points on a perpendicular bisector being equidistant from the endpoints of a segment, which is not relevant here.

Answer:

SSS congruence theorem