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} = $	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 \(\frac{CB}{DE}\)

Since \(\triangle CAB \sim \triangle DAE\) (from Part C), the ratio of corresponding sides \( \frac{CB}{DE} \) should be equal to the ratio of other corresponding sides, like \( \frac{CA}{DA} \). From Part A, we know \( \frac{CA}{DA} = 2 \), so \( \frac{CB}{DE} = \frac{CA}{DA} = 2 \) (by the property of similar triangles, corresponding sides are in proportion). So the ratio in step 12 is \( \frac{CA}{DA} \) (or equivalently, from Part A, we can use the ratio we found there). Wait, actually, in step 12, the ratio \( \frac{CB}{DE} \) is equal to the ratio of corresponding sides, which from Part A we know \( \frac{CA}{DA} = 2 \), and also from the similar triangles, the ratio of \( CB \) to \( DE \) should be equal to the ratio of \( CA \) to \( DA \) (since they are corresponding sides). So step 12: \( \frac{CB}{DE} = \frac{CA}{DA} \) (because in similar triangles \(\triangle CAB\) and \(\triangle DAE\), \( CB \) corresponds to \( DE \), and \( CA \) corresponds to \( DA \)). Then step 13: substitute \( \frac{CA}{DA} = 2 \) (from Part A) into \( \frac{CB}{DE} \), so \( \frac{CB}{DE} = 2 \). Then step 14: to solve for \( DE \), multiply both sides of \( \frac{CB}{DE} = 2 \) by \( DE \) and then divide by 2, or more directly, multiply both sides by \( \frac{DE}{2} \), but actually, starting from \( \frac{CB}{DE} = 2 \), we can rearrange to \( CB = 2 \cdot DE \), then divide both sides by 2 to get \( DE = \frac{1}{2}CB \). So for step 12, the ratio is \( \frac{CA}{DA} \) (which is 2), step 13: substitute the value from Part A, so \( \frac{CB}{DE} = 2 \), and step 14: multiply both sides by \( \frac{DE}{2} \) or divide both sides by 2 and multiply by \( DE \), but the operation is to solve \( \frac{CB}{DE} = 2 \) for \( DE \), so we can multiply both sides by \( DE \) to get \( CB = 2 \cdot DE \), then divide both sides by 2: \( DE = \frac{1}{2}CB \). So the expression to multiply both sides by is \( \frac{DE}{2} \)? Wait, no, let's do it step by step.

Step 12: Since \(\triangle CAB \sim \triangle DAE\), the ratio of corresponding sides \( \frac{CB}{DE} = \frac{CA}{DA} \) (by definition of similar triangles, corresponding sides are proportional). So the "Pick ratio" in step 12 is \( \frac{CA}{DA} \).

Step 13: Substitute \( \frac{CA}{DA} = 2 \) (from Part A) into \( \frac{CB}{DE} = \frac{CA}{DA} \), so \( \frac{CB}{DE} = 2 \).

Step 14: To solve \( \frac{CB}{DE} = 2 \) for \( DE \), we can multiply both sides by \( DE \) to get \( CB = 2 \cdot DE \), then divide both sides by 2: \( DE = \frac{1}{2}CB \). Alternatively, we can multiply both sides by \( \frac{DE}{2} \), but the operation is multiplying both sides by \( \frac{DE}{2} \) would be incorrect. Wait, actually, starting from \( \frac{CB}{DE} = 2 \), multiply both sides by \( DE \): \( CB = 2 \cdot DE \), then divide both sides by 2: \( \frac{CB}{2} = DE \), which is \( DE = \frac{1}{2}CB \). So the "Pick expression" in step 14 is \( \frac{DE}{2} \)? No, the operation is to multiply both sides by \( \frac{DE}{2} \)? Wait, no, let's see:

We have \( \frac{CB}{DE} = 2 \).

Multiply both sides by \( DE \): \( CB = 2 \cdot DE \).

Then divide both sides by 2: \( \frac{CB}{2} = DE \), which is \( DE = \frac{1}{2}CB \).

So the operation in step 14 is multiplying both sides by \( \frac{DE}{2} \)? No, actually, the "multiply both sides by" part: we can think of it as multiplying both sides by \( \frac{DE}{2} \), but maybe more clearly, from \( \frac{CB}{DE} = 2 \), we can solve for \( DE \) by cross - multiplying or rearranging.

But let's fill in the blanks:

Step 12: \( \frac{CB}{DE} = \frac{CA}{DA} \) (because in similar triangles, corresponding sides are proportional; \( CB \) corresponds to \( DE \), \( CA \) corresponds to \( DA \))

Step 13: Substitute \( \frac{CA}{DA} = 2 \) (from Part A) into the equation from step 12, so \( \frac{CB}{DE} = 2 \)

Step 14: To get from \( \frac{CB}{DE} = 2 \) to \( \frac{1}{2}CB = DE \), we can multiply both sides of \( \frac{CB}{DE} = 2 \) by \( \frac{DE}{2} \). Let's check:

Left - hand side: \( \frac{CB}{DE} \times \frac{DE}{2}=\frac{CB}{2} \)

Right - hand side: \( 2\times\frac{DE}{2}=DE \)

So we multiply both sides by \( \frac{DE}{2} \)

Step 1: Step 12 - Ratio of corresponding sides

In similar triangles \(\triangle CAB\) and \(\triangle DAE\), the ratio of corresponding sides \( \frac{CB}{DE} \) is equal to the ratio of other corresponding sides, such as \( \frac{CA}{DA} \). So \( \frac{CB}{DE}=\frac{CA}{DA} \) (by the property of similar triangles: corresponding sides are in proportion).

Step 2: Step 13 - Substitution

From Part A, we know that \( \frac{CA}{DA} = 2 \). Substitute \( \frac{CA}{DA} = 2 \) into the equation \( \frac{CB}{DE}=\frac{CA}{DA} \) (from step 12). So \( \frac{CB}{DE}=2 \).

Step 3: Step 14 - Solving for \( DE \)

We start with the equation \( \frac{CB}{DE}=2 \). To solve for \( DE \), we multiply both sides of the equation by \( \frac{DE}{2} \).

Left - hand side: \( \frac{CB}{DE}\times\frac{DE}{2}=\frac{CB}{2} \)

Right - hand side: \( 2\times\frac{DE}{2}=DE \)

So we get \( \frac{1}{2}CB = DE \), or \( DE=\frac{1}{2}CB \)

Answer:

Step 12: \(\frac{CA}{DA}\) (or the ratio from the similar triangle's corresponding sides, which is equal to 2 from Part A)
Step 13: \(2\)
Step 14: \(\frac{DE}{2}\) (the expression to multiply both sides by)

And the final proof shows that \(DE = \frac{1}{2}CB\) by using the properties of similar triangles and substitution. The filled - in blanks are:

1. \(\frac{CA}{DA}\)
2. \(2\)
3. \(\frac{DE}{2}\) (and then we get \(DE=\frac{1}{2}CB\))