On the left constructed triangle FED with right angle D. On the right, triangle ABC with AC = b, CB = a, and AB = c.

Question

Given: ΔABC where CB=a , AC=b , and AB=c and a2+b2=c2

Construct: Right ΔDEF such that AC¯¯¯¯¯¯¯¯≅FD¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯≅DE¯¯¯¯¯¯¯¯

Prove: ΔABC is a right triangle

Read the statements of proof. Then, answer the question.

Statements Reasons
1. ΔABC where CB=a , AC=b , and AB=c and a2+b2=c2 1. Given
2. Construct right ΔDEF such that AC¯¯¯¯¯¯¯¯≅FD¯¯¯¯¯¯¯¯ and CB¯¯¯¯¯¯¯¯≅DE¯¯¯¯¯¯¯¯ 2. Construction
3. (AB)2=(AC)2+(CB)2 3. Substitution Property
4. 4. Pythagorean Theorem
5. AC=FD

CB=DE

5. Definition of congruence
6. (EF)2=(AC)2+(CB)2 6.
7. (EF)2=(AB)2 7. Transitive Property
8. 8. Calculations (square root)
9. EF¯¯¯¯¯¯¯¯≅AB¯¯¯¯¯¯¯¯ 9. Definition of congruence
10. ΔABC≅ΔFED 10. SSS Congruence Theorem
11. ∠C≅∠D 11. CPCTC Theorem
12. m∠C=m∠D 12. Definition of congruence
13. m∠D=90° 13.
14. m∠C=90° 14.
15. ΔABC is a right triangle 15. Definition of right triangles
What statement belongs in line 4?

(1 point)
Responses

(EF)2=(FD)2−(DE)2
cap e cap f squared is equal to cap f cap d squared minus cap d cap e squared

(FD)2=(EF)2+(DE)2
cap f cap d squared is equal to cap e cap f squared plus cap d cap e squared

(EF)2=(FD)2+(DE)2
cap e cap f squared is equal to cap f cap d squared plus cap d cap e squared

(DE)2=(FD)2−(DE)2

Answer 1

To determine the correct statement for line 4, we need to connect the properties of right triangles and the construction we have made.

In triangle \(DEF\), which is a right triangle with \(\angle D = 90^\circ\), we can apply the Pythagorean Theorem. According to the Pythagorean Theorem:

\[ (DF)^2 = (DE)^2 + (EF)^2 \]

However, since line 4 should relate to what we already have established, it should describe the sum of the squares of the legs of triangle \(DEF\) equating to the square of the hypotenuse.

Taking into account how the sides are labeled and defined:

• \(FD\) serves as the hypotenuse
• \(DE\) and \(EF\) serve as the legs

So we can state in line 4:

\[ (FD)^2 = (DE)^2 + (EF)^2 \]

This matches option:

• (FD)² = (EF)² + (DE)²

This notation correctly aligns with the Pythagorean theorem applied to triangle \(DEF\).

Therefore, the correct line for statement 4 is:

(FD)² = (EF)² + (DE)².