Just give answers to blanck ones just give me The answers cause I don't want no explaining because I'm Savage

statuscvo [17]

Problem 4, part (a)

<h3>Answer: $\triangle\text{L}\text{B}\text{M}$ </h3>

Explanation:

Notice that $\frac{\text{A}\text{B}}{\text{L}\text{B}} = \frac{60}{24} = 2.5$ and $\frac{\text{B}\text{C}}{\text{B}\text{M}} = \frac{32+48}{32} = 2.5$ ; both ratios are equal to 2.5

The two triangles have the common overlapped or shared angle at $\text{A}\text{B}\text{C}$ , which is identical to angle $\text{L}\text{B}\text{M}$ .

Therefore, we can use the SAS similarity theorem to prove triangle $\text{A}\text{B}\text{C}$ is similar to triangle $\text{L}\text{B}\text{M}$ .

===========================================

Problem 4, part (b)

<h3>Answer: AC and LM</h3>

Explanation:

Similar triangles have congruent corresponding angles.

Since $\triangle ABC \sim \triangle LBM$ , we know that $\angle CAB \cong \angle MLB$ . These corresponding angles then lead to AC being parallel to LM. Refer to the converse of the corresponding angles theorem.

===========================================

Problem 4, part (c)

If we want to prove that the triangles are all similar using SSS, then we need all three of the following statements to be true

$\frac{\text{A}\text{B}}{\text{N}\text{M}} = 2.5$

$\frac{\text{B}\text{C}}{\text{M}\text{C}} = 2.5$

$\frac{\text{A}\text{C}}{\text{N}\text{C}} = 2.5$

Unfortunately, the reality is that $\frac{\text{A}\text{B}}{\text{N}\text{M}} = \frac{60}{35} \approx 1.71$ doesn't match with the 2.5; so the three triangles are definitely not similar. We need to change NM = 35 to NM = 24 so that we have similar triangles. We just copy what segment LB shows.

------------

If instead you wanted to use SAS, then we would need NM = 24 like earlier. Also, we would need angle ABC = angle NMC to be true. Lastly, we need MC = 32 so it matches up with MB = 32.

------------

If you want to use the AA similarity rule, then we need these statements below to be true

$\angle \text{A}\text{B}\text{C} \cong\angle \text{N}\text{M}\text{C}\\\angle \text{A}\text{C}\text{B} \cong\angle \text{N}\text{C}\text{M}\\\angle \text{C}\text{A}\text{B} \cong\angle \text{C}\text{N}\text{M}\\$

As you can see, there are few pathways we can take to prove the triangles similar.

===========================================

Problem 5

<h3>Refer to the screenshot below. </h3>

I've filled out the table with the correct items.

You are correct to start with the given statement, which is how <u>all</u> proofs start off.

On the complete opposite end of the spectrum, the last statement will be what we want to prove. Which is namely that $\triangle \text{A}\text{C}\text{E} \sim \triangle \text{B}\text{C}\text{D}$ , i.e. that those triangles are similar.

So somehow we have to connect the given to the thing we want to prove at the end.

Notice that angles $\text{C}\text{B}\text{D}$ and $\text{C}\text{A}\text{E}$ are corresponding angles. They are congruent because of the parallel lines. So we'll have $\angle \text{C}\text{B}\text{D} \cong \angle \text{C}\text{A}\text{E}$ for statement 2. Using identical logic, we will also have $\angle \text{C}\text{D}\text{B} = \angle \text{C}\text{E}\text{A}$ for statement 3. Both statements 2 and 3 use the reasoning of "corresponding angles are congruent". Keep in mind that the statement in quotes is only true when we have parallel lines like this.

Lastly, we'll use the AA similarity theorem to fully prove what we want, which is that $\triangle \text{A}\text{C}\text{E}$ is similar to triangle $\triangle\text{B}\text{C}\text{D}$ .