Question

modeling real life the navajo rug is made of isosceles triangles. you know $\angle b \cong \angle d$. use the sas congruence theorem to prove that $\triangle abc \cong \triangle cde$.
because $\triangle abc$, $\triangle bcd$, and $\triangle cde$ are isosceles triangles, you know that $\overline{ab} \cong \boxed{\quad}$, $\overline{bc} \cong \boxed{\quad}$, and $\overline{cd} \cong \boxed{\quad}$.
so by the \boxed{\quad} $\overline{ab} \cong \overline{cd}$ and $\overline{bc} \cong \overline{de}$.
its given that $\angle b \cong \boxed{\quad}$, so $\triangle abc \cong \triangle cde$ by the sas congruence theorem.

options for congruence segments: $\overline{ac}$, $\overline{bc}$, $\overline{cd}$, $\overline{ce}$, $\overline{de}$
options for congruence property: reflexive property of congruence, symmetric property of congruence, transitive property of congruence

Explanation:

Step1: Analyze SAS for $\triangle ABC \cong \triangle CDE$

SAS (Side - Angle - Side) requires two sides and the included angle. We know $\angle B \cong \angle D$ (given). For sides: since $\triangle ABC$, $\triangle BCD$, $\triangle CDE$ are isosceles, $AB = BC$ (in $\triangle ABC$), $BC = CD$ (in $\triangle BCD$), $CD = DE$ (in $\triangle CDE$). By Transitive Property of Congruence, $AB \cong CD$ and $BC \cong DE$.

Step2: Fill in the blanks for congruent sides and angle

• First blank (for $AB \cong$): From above, $AB \cong CD$.
• Second blank (for $BC \cong$): $BC \cong DE$.
• Third blank (for $CD \cong$): $CD \cong AB$ (Symmetric Property, but from isosceles and transitive, also $CD \cong AB$ as we saw $AB = CD$). Wait, no, the angle is $\angle B \cong \angle D$. Wait, the SAS: sides $AB$ and $BC$ in $\triangle ABC$, sides $CD$ and $DE$ in $\triangle CDE$, and included angle $\angle B$ and $\angle D$. So:
• $AB \cong CD$ (first blank: $CD$)
• $BC \cong DE$ (second blank: $DE$)
• $CD \cong AB$ (but the third blank for $CD \cong$? Wait, no, the problem says "you know that $\overline{AB} \cong \square$, $\overline{BC} \cong \square$, and $\overline{CD} \cong \square$". Wait, no, re - reading: "Because $\triangle ABC$, $\triangle BCD$, and $\triangle CDE$ are isosceles triangles, you know that $\overline{AB} \cong \square$, $\overline{BC} \cong \square$, and $\overline{CD} \cong \square$".
• In isosceles $\triangle ABC$: $AB = BC$? No, wait, isosceles triangle has two equal sides. Maybe $\triangle ABC$ has $AB = AC$? No, the diagram (Navajo rug with isosceles triangles) – probably $\triangle ABC$: $AB = BC$, $\triangle BCD$: $BC = CD$, $\triangle CDE$: $CD = DE$. So by Transitive, $AB = CD$, $BC = DE$, $CD = AB$ (Symmetric) or $CD = DE$? Wait, no, let's correct:
• $\triangle ABC$ isosceles: let's say $AB = BC$
• $\triangle BCD$ isosceles: $BC = CD$
• $\triangle CDE$ isosceles: $CD = DE$
• So $AB \cong CD$ (Transitive: $AB = BC = CD$)
• $BC \cong DE$ (Transitive: $BC = CD = DE$)
• $CD \cong AB$ (Symmetric Property, since $AB \cong CD$)
• Then the angle: $\angle B \cong \angle D$ (given, as $\angle B$ is in $\triangle ABC$, $\angle D$ in $\triangle CDE$)
• Then "It's given that $\angle B \cong \angle D$, so $\triangle ABC \cong \triangle CDE$ by the SAS Congruence Theorem."
• For the property used for $AB \cong CD$ and $BC \cong DE$: Transitive Property of Congruence (since $AB = BC$, $BC = CD$ implies $AB = CD$; $BC = CD$, $CD = DE$ implies $BC = DE$)

Answer:

• First blank (for $\overline{AB} \cong$): $\overline{CD}$
• Second blank (for $\overline{BC} \cong$): $\overline{DE}$
• Third blank (for $\overline{CD} \cong$): $\overline{AB}$ (or $\overline{DE}$? Wait, no, $CD = DE$ from $\triangle CDE$ being isosceles. Wait, I think I made a mistake. Let's start over.

Correct approach:

1. Isosceles triangle $\triangle ABC$: let the equal sides be $AB = BC$ (so $\overline{AB} \cong \overline{BC}$)
2. Isosceles triangle $\triangle BCD$: $BC = CD$ (so $\overline{BC} \cong \overline{CD}$)
3. Isosceles triangle $\triangle CDE$: $CD = DE$ (so $\overline{CD} \cong \overline{DE}$)

By Transitive Property of Congruence:

• $\overline{AB} \cong \overline{CD}$ (because $\overline{AB} \cong \overline{BC}$ and $\overline{BC} \cong \overline{CD}$)
• $\overline{BC} \cong \overline{DE}$ (because $\overline{BC} \cong \overline{CD}$ and $\overline{CD} \cong \overline{DE}$)
• $\overline{CD} \cong \overline{AB}$ (Symmetric Property, since $\overline{AB} \cong \overline{CD}$) or $\overline{CD} \cong \overline{DE}$ (from $\triangle CDE$)

But for SAS:

In $\triangle ABC$ and $\triangle CDE$:

• Side 1: $\overline{AB}$ and $\overline{CD}$ (congruent, from transitive)
• Side 2: $\overline{BC}$ and $\overline{DE}$ (congruent, from transitive)
• Included angle: $\angle B$ and $\angle D$ (given $\angle B \cong \angle D$)

So the blanks:

• $\overline{AB} \cong \overline{CD}$
• $\overline{BC} \cong \overline{DE}$
• $\overline{CD} \cong \overline{AB}$ (or $\overline{DE}$? No, $\overline{CD} \cong \overline{AB}$ because $\overline{AB} \cong \overline{CD}$ (transitive) and symmetric property)

And the angle: $\angle B \cong \angle D$ (given), so $\triangle ABC \cong \triangle CDE$ by SAS.

For the property used for the congruence of sides (the "So by the $\square$" part): Transitive Property of Congruence (since we used transitive to get $AB \cong CD$ and $BC \cong DE$ from the isosceles triangle equal sides).

Final blanks:

• $\overline{AB} \cong \overline{CD}$
• $\overline{BC} \cong \overline{DE}$
• $\overline{CD} \cong \overline{AB}$ (or $\overline{DE}$? Wait, no, $\overline{CD} \cong \overline{DE}$ from $\triangle CDE$ being isosceles. I think I confused the triangles. Let's look at the options for the property: Reflexive, Symmetric, Transitive.

The sides: $AB = BC$ (Reflexive? No, Reflexive is $AB = AB$). Symmetric: if $AB = CD$, then $CD = AB$. Transitive: if $AB = BC$ and $BC = CD$, then $AB = CD$. So the property is Transitive Property of Congruence.

So:

• $\overline{AB} \cong \overline{CD}$ (because $AB = BC$, $BC = CD$; transitive)
• $\overline{BC} \cong \overline{DE}$ (because $BC = CD$, $CD = DE$; transitive)
• $\overline{CD} \cong \overline{AB}$ (symmetric, since $AB = CD$) or $\overline{CD} \cong \overline{DE}$ (from $\triangle CDE$)

But the problem is to fill in the blanks as per the SAS to prove $\triangle ABC \cong \triangle CDE$.

SAS requires two sides and the included angle. So:

• Side 1: $AB$ and $BC$ in $\triangle ABC$
• Side 2: $CD$ and $DE$ in $\triangle CDE$
• Included angle: $\angle B$ and $\angle D$

So:

• $\overline{AB} \cong \overline{CD}$ (first blank: $CD$)
• $\overline{BC} \cong \overline{DE}$ (second blank: $DE$)
• $\overline{CD} \cong \overline{AB}$ (third blank: $AB$)
• Angle: $\angle B \cong \angle D$ (given)
• Property for side congruence: Transitive Property of Congruence