Question

below is the proof that $de = \frac{1}{2}cb$. the proof is divided into four parts, where the title of each part indicates its main purpose. complete part d of the proof. part a: prove $\frac{ca}{da} = 2$ show the steps. part b: prove $\frac{ba}{ea} = 2$ show the steps. part c: prove $\triangle cab \sim \triangle dae$ show the steps. part d: prove $de = \frac{1}{2}cb$ statement reason 12 $\frac{cb}{de} = $ pick ratio lengths of corresponding sides of similar triangles have equal ratios. (part c) 13 $\frac{cb}{de} = \square$ substitution (part a, 12) 14 $\frac{1}{2}cb = de$ multiply both sides of the equation by pick expression. (13)

Explanation:

Step 1: Determine the ratio for similar triangles

In similar triangles \( \triangle CAB \sim \triangle DAE \), the ratio of corresponding sides is equal. From Part A, we know \( \frac{CA}{DA} = 2 \), and since the triangles are similar, the ratio of \( CB \) to \( DE \) should be equal to the ratio of \( CA \) to \( DA \) (or \( BA \) to \( EA \)). So \( \frac{CB}{DE}=\frac{CA}{DA} \) (or \( \frac{BA}{EA} \), but since \( \frac{CA}{DA} = 2 \) from Part A, we use that).

Step 2: Substitute the known ratio

We know from Part A that \( \frac{CA}{DA}=2 \), so substituting into \( \frac{CB}{DE}=\frac{CA}{DA} \), we get \( \frac{CB}{DE} = 2 \).

Step 3: Solve for \( DE \)

To solve for \( DE \), we multiply both sides of the equation \( \frac{CB}{DE}=2 \) by \( \frac{DE}{2} \) (or equivalently, multiply both sides by \( DE \) and then divide by 2, but the operation is multiplying both sides by \( \frac{DE}{2} \) to isolate \( DE \)). So starting from \( \frac{CB}{DE}=2 \), multiply both sides by \( \frac{DE}{2} \): \( CB = 2DE \), then divide both sides by 2 to get \( DE=\frac{1}{2}CB \). For the table:

• Step 12: The ratio \( \frac{CB}{DE}=\frac{CA}{DA} \) (or \( \frac{BA}{EA} \), but since \( \frac{CA}{DA}=2 \), we can use \( \frac{CA}{DA} \) which is 2, so the ratio is \( \frac{CA}{DA} \) (or 2, but the ratio from similar triangles is \( \frac{CB}{DE}=\frac{CA}{DA} \))
• Step 13: Substitute \( \frac{CA}{DA}=2 \) into \( \frac{CB}{DE}=\frac{CA}{DA} \), so \( \frac{CB}{DE}=2 \)
• Step 14: Multiply both sides of \( \frac{CB}{DE}=2 \) by \( \frac{DE}{2} \) (the expression is \( \frac{DE}{2} \)) to get \( DE = \frac{1}{2}CB \)

For the table entries:

• Step 12: The ratio is \( \frac{CA}{DA} \) (or \( \frac{BA}{EA} \), but since \( \frac{CA}{DA}=2 \), we use that ratio, so \( \frac{CB}{DE}=\frac{CA}{DA} \))
• Step 13: \( \frac{CB}{DE}=2 \)
• Step 14: The expression to multiply both sides by is \( \frac{DE}{2} \) (or equivalently, we can think of it as multiplying both sides by \( DE \) and then dividing by 2, but the operation is multiplying by \( \frac{DE}{2} \))

Answer:

• Step 12: \( \frac{CA}{DA} \) (or \( \frac{BA}{EA} \), but using the ratio from Part A, \( \frac{CA}{DA} \))
• Step 13: \( 2 \)
• Step 14: \( \frac{DE}{2} \) (the expression to multiply both sides by is \( \frac{DE}{2} \) to get \( DE=\frac{1}{2}CB \))

For the table in the problem:

1. \( \frac{CB}{DE}=\boldsymbol{\frac{CA}{DA}} \) (or \( \frac{BA}{EA} \), but \( \frac{CA}{DA}=2 \))
2. \( \frac{CB}{DE}=\boldsymbol{2} \)
3. Multiply both sides by \( \boldsymbol{\frac{DE}{2}} \)