# Is there an analog to the SSS triangle congruence theorem for quadrilaterals?

Question
Congruence
Is there an analog to the SSS triangle congruence theorem for quadrilaterals?

2020-10-24
Step 1
To show whether there is an analog to the SSS triangle congruence theorem for quadrilateral.
Step 2
The SSS triangle congruence theorem states that if all 3 sides of one triangle are in proportion to all 3 sides of another triangle then those triangles are similar.
Step 3
It is observed that Side-Side-Side congruence is not sufficient to prove that two quadrilaterals are congruent. Since there are 4 parameters for a quadrilateral SSSS is possible.
Step 4
Therefore, there is no analog to the SSS triangle congruence theorem for quadrilateral.

### Relevant Questions

With of the following triangle congruence shortcuts could be used to prove PRQ = TRS

Given Data,
$$\displaystyle\angle{Q}\stackrel{\sim}{=}\angle{S}$$
$$\displaystyle\overline{{{Q}{R}}}\stackrel{\sim}{=}\overline{{{S}{R}}}$$
a)Side-Side-Side Postulate (SSS)
b)Side-Angle-Side Postulate (SAS)
c)Angle-Side-Angle Postulate (ASA)
d)Angle-Angle-Side Theorem (AAS)
Select all statements that are true about the triangles.

-Triangle ABC and DCB are congruent by the Angle-Angle Triangle Congruence theorem.
-Triangle ABC and BCD are congruent by the Angle-Side-Angle Triangle Congruence theorem.
-Triangle ABC and BCD are congruent by the Side-Side-Side Triangle Congruence theorem.
-Triangle ABC and DCB are congruent by the Side-Angle-Side Triangle Congruence theorem.
-Triangle ABC and DCB are congruent by the Side-Side-Side Triangle Congruence theorem.
-There is not enough information to determine if the triangles are congruent.
Solve the following congruence.
$$\displaystyle{x}^{{{2}}}\equiv{7}\pm{o}{d}{29}$$
Find the triangle congruence (SSS SAS AAS HL)

Given: bar(SQ) and bar(PR) bisect each other.
To determine:The check digit of an ISSN that has these initial 7 digits 1553-734
In congruence classes $$\displaystyle\frac{{Z}}{{{m}{Z}}}$$, reduce the equation $$\displaystyle{a}_{{m}}\cdot{{x}_{{m}}^{{2}}}={c}_{{m}}$$ either by finding convenient representation for $$\displaystyle{a}_{{m}}{\quad\text{and}\quad}{b}_{{m}}$$ or by using the inverse of $$\displaystyle{a}_{{m}}$$. Then find a solution for this congruence directly or by replacing $$\displaystyle{c}_{{m}}$$ : with its appropriate representative in $$\displaystyle\frac{{Z}}{{{m}{Z}}}$$. If there is no solution explain why. Here $$\displaystyle{a}_{{m}},{b}_{{m}},{x}_{{m}}{\left(={x}\right)},{c}_{{m}}\in\frac{{Z}}{{{m}{Z}}}:$$
$$\displaystyle{I}{n}\frac{{Z}}{{{19}{Z}}},{\left[{2}\right]}\cdot{x}^{{2}}={\left[{13}\right]}:$$