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Sagot :
To determine which statement completes this proof, let's analyze the given steps in detail:
1. [tex]$\overline{CD}$[/tex] is an altitude of [tex]$\triangle ABC$[/tex].
Reason: Given.
2. [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles.
Reason: Definition of altitude (an altitude creates a right angle with the base of the triangle).
3. [tex]$\triangle ADC$[/tex] and [tex]$\triangle BCD$[/tex] are right triangles.
Reason: Both angles [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles, thus forming right triangles.
4. [tex]$\sin(A) = \frac{CD}{b}$[/tex] and [tex]$\sin(B) = \frac{CD}{a}$[/tex].
Reason: Definition of sine in right triangles:
- [tex]$\sin(A)$[/tex] is the ratio of the length of the side opposite angle A ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle ADC$[/tex] ([tex]$b$[/tex]).
- Similarly, [tex]$\sin(B)$[/tex] is the ratio of the length of the side opposite angle B ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle BCD$[/tex] ([tex]$a$[/tex]).
5. Multiplication property of equality:
From the equations in step 4, we can derive:
- Multiplying both sides of [tex]$\sin(A) = \frac{CD}{b}$[/tex] by [tex]$b$[/tex], we get [tex]$b \cdot \sin(A) = CD$[/tex].
- Multiplying both sides of [tex]$\sin(B) = \frac{CD}{a}$[/tex] by [tex]$a$[/tex], we get [tex]$a \cdot \sin(B) = CD$[/tex].
6. [tex]$CD = b \cdot \sin(A)$[/tex] and [tex]$CD = a \cdot \sin(B)$[/tex].
Reason: From the multiplication property of equality, we get these two relations.
7. By the Substitution property of equality, we equate the two expressions for [tex]$CD$[/tex]:
[tex]$b \cdot \sin(A) = a \cdot \sin(B)$[/tex].
- Dividing both sides by [tex]$\sin(A) \cdot \sin(B)$[/tex], we get [tex]$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$[/tex].
8. Finally, we conclude that:
- [tex]$\boxed{b = CD \cdot \sin(A) \text{ and } a = CD \cdot \sin(B)}$[/tex].
So, the correct statement from the provided options to complete the proof is:
C. [tex]$\quad b = CD \sin (A)$[/tex] and [tex]$a = CD \sin (B)$[/tex].
1. [tex]$\overline{CD}$[/tex] is an altitude of [tex]$\triangle ABC$[/tex].
Reason: Given.
2. [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles.
Reason: Definition of altitude (an altitude creates a right angle with the base of the triangle).
3. [tex]$\triangle ADC$[/tex] and [tex]$\triangle BCD$[/tex] are right triangles.
Reason: Both angles [tex]$\angle ADC$[/tex] and [tex]$\angle BDC$[/tex] are right angles, thus forming right triangles.
4. [tex]$\sin(A) = \frac{CD}{b}$[/tex] and [tex]$\sin(B) = \frac{CD}{a}$[/tex].
Reason: Definition of sine in right triangles:
- [tex]$\sin(A)$[/tex] is the ratio of the length of the side opposite angle A ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle ADC$[/tex] ([tex]$b$[/tex]).
- Similarly, [tex]$\sin(B)$[/tex] is the ratio of the length of the side opposite angle B ([tex]$CD$[/tex]) to the hypotenuse of triangle [tex]$\triangle BCD$[/tex] ([tex]$a$[/tex]).
5. Multiplication property of equality:
From the equations in step 4, we can derive:
- Multiplying both sides of [tex]$\sin(A) = \frac{CD}{b}$[/tex] by [tex]$b$[/tex], we get [tex]$b \cdot \sin(A) = CD$[/tex].
- Multiplying both sides of [tex]$\sin(B) = \frac{CD}{a}$[/tex] by [tex]$a$[/tex], we get [tex]$a \cdot \sin(B) = CD$[/tex].
6. [tex]$CD = b \cdot \sin(A)$[/tex] and [tex]$CD = a \cdot \sin(B)$[/tex].
Reason: From the multiplication property of equality, we get these two relations.
7. By the Substitution property of equality, we equate the two expressions for [tex]$CD$[/tex]:
[tex]$b \cdot \sin(A) = a \cdot \sin(B)$[/tex].
- Dividing both sides by [tex]$\sin(A) \cdot \sin(B)$[/tex], we get [tex]$\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$[/tex].
8. Finally, we conclude that:
- [tex]$\boxed{b = CD \cdot \sin(A) \text{ and } a = CD \cdot \sin(B)}$[/tex].
So, the correct statement from the provided options to complete the proof is:
C. [tex]$\quad b = CD \sin (A)$[/tex] and [tex]$a = CD \sin (B)$[/tex].
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