Geometry assignment worksheet focusing on triangle similarity using AA, SAS, and SSS criteria.

Let’s go through each problem one by one. We’re checking if the triangles are similar, and if so, which rule proves it: AA (Angle-Angle), SAS (Side-Angle-Side), or SSS (Side-Side-Side).

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Problem 1)
We have two triangles: △EFJ and △KJL (they share point J).
Look at the angles:
- Angle F is marked with a red arc → same as angle K (also marked with red arc) → so ∠F ∠K
- Also, angles at J are vertical angles → they are equal → ∠EJF ≅ ∠KJL

So we have two pairs of congruent angles → that’s AA similarity.

✔ Answer: C) similar; AA similarity

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Problem 2)
Triangles: △QCP and △ECD (they cross at point C).
Given:
- ∠Q = 68°
- ∠D = 68° → so ∠Q ≅ ∠D
- Angles at C are vertical angles → ∠QCP ≅ ∠ECD

Again, two pairs of congruent angles → AA similarity

✔ Answer: B) similar; AA similarity

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Problem 3)
Triangle MLK has a smaller triangle UT inside? Wait — actually, points U and T are on sides MK and LK respectively.
Given angles:
- In big triangle MLK: ∠M = 77°, ∠K = 41° → so ∠L = 180 - 77 - 41 = 62°
- In small triangle UTK: ∠U = 77°, ∠K = 41° → so ∠T = 62° too!

Wait — but are these the same triangles? Actually, looking at diagram:
△MLK and △UTK share angle K (41°), and both have a 77° angle (∠M and ∠U). So again, two angles match → AA similarity

But wait — let me double-check labels. The small triangle is △UTK? Or △UT something? Diagram shows:
Points: M-U-K on top, L-T-K on bottom. So triangle MLK and triangle UTK? Yes.
∠M = 77°, ∠U = 77° → corresponding? If U is on MK and T is on LK, then yes — ∠M corresponds to ∠U, and ∠K is shared. So AA.

✔ Answer: A) similar; AA similarity

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Problem 4)
Triangle UST with line NM parallel to ST? Not stated, but given angles:
In triangle UNS: ∠N = 73°, ∠S = 71° → so ∠U = 180 - 73 - 71 = 36°
In big triangle UST: we don’t know other angles yet. But if NM is drawn such that it creates triangle UNM and trapezoid NMTS... Wait, actually, the figure shows triangle UST with points N on US and M on UT, and segment NM. Given ∠UNM = 73°, ∠UST = 71°.

Actually, better approach:
If we assume NM is not necessarily parallel, but we can compute angles.

In triangle UNS: angles are 73° at N, 71° at S → so angle at U is 36°.

Now look at triangle UST: we only know angle at S is 71°. Do we know angle at U? It’s the same vertex — so angle at U in big triangle is also 36°? Only if N and M are on the sides and we’re comparing △UNM and △UST.

Wait — perhaps the intended comparison is between △UNM and △UST.

In △UNM: ∠U = ? , ∠N = 73°, ∠M = ?

Actually, maybe I misread. Let me re-express:

The diagram likely shows triangle UST, with points N on US and M on UT, and segment NM. Given: ∠UNM = 73°, ∠UST = 71°.

But without more info, we can’t say for sure unless we assume something.

Wait — another way: if we consider triangle UNS and triangle UST — no.

Perhaps the key is: in triangle UNS, angles are 73°, 71°, so third angle is 36°.

In triangle UST, if we assume that point M is on UT and N on US, and we’re comparing △UNM and △UST — but we don’t have enough angles for △UNM.

Wait — maybe the 73° is at N in triangle UNM, and 71° is at S in triangle UST. That doesn't directly help.

Alternatively, perhaps the figure implies that NM is parallel to ST? Because often in such problems, if a line cuts two sides and forms equal corresponding angles, it's parallel.

If ∠UNM = 73° and ∠UST = 71°, those are not equal, so probably not parallel.

Wait — let’s calculate all angles.

Assume we are comparing △UNM and △UST.

In △UNM: we know ∠UNM = 73°. What about ∠NUM? That’s part of angle at U.

This is getting messy. Maybe there’s a better way.

Looking back: in many textbooks, when you have a triangle with a line connecting two sides, and you’re given two angles, you can find if the small triangle is similar to the big one.

But here, in △UNS (if that’s the small triangle), angles are 73° at N, 71° at S → so angle at U is 36°.

In big triangle UST, angle at S is 71°, angle at U is the same 36° (since it’s the same vertex), so angle at T would be 180 - 71 - 36 = 73°.

Oh! So in big triangle UST: ∠U=36°, ∠S=71°, ∠T=73°

In small triangle UNS: ∠U=36°, ∠N=73°, ∠S=71° — same angles!

So △UNS ~ △UST? But that doesn’t make sense because UNS is part of UST.

Actually, the small triangle should be △UNM or something else.

I think I made a mistake in labeling.

Let me reinterpret: the diagram probably shows triangle UST, with points N on US and M on UT, and segment NM. Given: ∠UNM = 73°, and ∠UST = 71°.

But ∠UNM is an angle in triangle UNM, while ∠UST is in triangle UST.

To compare △UNM and △UST:

- They share angle at U.
- If we can show another angle equal, then AA.

But we have ∠UNM = 73°, and in △UST, what is the corresponding angle? If NM were parallel to ST, then ∠UNM would correspond to ∠UST, but 73° ≠ 71°, so not parallel.

Perhaps the 73° is meant to be at M or something.

Another idea: maybe the 73° is ∠UMN or ∠UNM — let's assume it's ∠UNM = 73°, and in the big triangle, ∠UTS might be 73°? But we don't know.

Perhaps calculate based on sum.

Let’s denote:

In triangle UNS (assuming S is connected): but point M is there.

I recall that in some versions of this problem, the angles are set so that the small triangle has angles matching the big one.

Let’s try this: suppose in triangle UST, we have points N on US, M on UT, and NM drawn. Given ∠UNM = 73°, and ∠UST = 71°.

Then in triangle UNS, if we consider it, but M is not in it.

Perhaps the small triangle is △UNM, and we need its angles.

We know ∠UNM = 73°. What is ∠NUM? It's the same as ∠SUT, which is common.

But we don't know it yet.

From the big triangle UST, if we knew two angles, but we only know ∠S = 71°.

Unless... perhaps the 73° is not in the small triangle but indicates something else.

I think there might be a standard interpretation. Let me search my memory.

Ah! Perhaps the 73° is the angle at N in triangle UNM, and since NM is a straight line, but no.

Another thought: in triangle UNS, if S is a vertex, but M is on UT, so triangle UNS includes points U,N,S, but S is connected to T, so it's not closed.

I think I found the issue: in the diagram, likely triangle MLK from problem 3 was clear, but for problem 4, it's triangle UST with N on US, M on UT, and NM drawn. Given ∠UNM = 73° and ∠UST = 71°.

Then, in triangle UNM, we have ∠UNM = 73°. The angle at U is common to both △UNM and △UST.

Let ∠U = x.

Then in △UNM, angles are: at U: x, at N: 73°, so at M: 180 - x - 73 = 107 - x.

In △UST, angles are: at U: x, at S: 71°, so at T: 180 - x - 71 = 109 - x.

For the triangles to be similar, corresponding angles must be equal.

Suppose △UNM ~ △UST. Then correspondence could be U->U, N->S, M->T.

Then ∠UNM should equal ∠UST, but 73° vs 71° — not equal.

Or U->U, N->T, M->S — then ∠UNM should equal ∠UTS, which is 109 - x, and we have 73 = 109 - x => x = 36.

Then check other angles: in △UNM, angle at M is 107 - x = 107 - 36 = 71°.

In △UST, angle at S is 71°, so if M corresponds to S, then ∠UMN = 71° = ∠UST, good.

And angle at U is 36° in both.

So yes! With x=36°, we have:

△UNM: angles 36° at U, 73° at N, 71° at M

△UST: angles 36° at U, 71° at S, 73° at T

So if we map U->U, N->T, M->S, then angles match: 36=36, 73=73, 71=71.

So they are similar by AA (actually AAA, but AA suffices).

So answer should be similar by AA similarity.

But let's confirm the correspondence: vertex U to U, N to T, M to S.

Yes, angles correspond.

So ✔ Answer: C) similar; AA similarity

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Problem 5)
Two separate triangles: △EFD and △TSU.

Sides of △EFD: EF=777, FD=666, ED=1147

Sides of △TSU: TS=358, SU=306, TU=527

Check if sides are proportional.

Compute ratios:

First, sort the sides to match smallest to largest.

△EFD: 666, 777, 1147

△TSU: 306, 358, 527

Now ratio of smallest: 666 / 306 ≈ ? Let's calculate.

666 ÷ 306 = 666/306 = 333/153 = 111/51 = 37/17 ≈ 2.176

Next: 777 / 358 ≈ ? 777 ÷ 358 ≈ 2.170 (since 358*2=716, 777-716=61, so 2 + 61/358 ≈ 2.170)

Last: 1147 / 527 ≈ ? 527*2=1054, 1147-1054=93, so 2 + 93/527 ≈ 2.176

All approximately 2.176, so yes, proportional.

Exactly: let's see if 666/306 = 777/358 = 1147/527

Simplify fractions:

666 and 306 divide by 18: 666÷18=37, 306÷18=17 → 37/17

777 and 358: 777÷? 358÷2=179, 777÷... gcd of 777 and 358.

358 factors: 2*179

777 ÷ 179? 179*4=716, 777-716=61, not divisible. 777 ÷ 3 = 259, 358 not div by 3.

Earlier calculation showed approx same.

1147 / 527: 527 * 2 = 1054, 1147 - 1054 = 93, so 2 + 93/527

93 and 527: 527 ÷ 17 = 31, since 17*31=527, 93÷31=3, so 93/527 = 3/17, so total 2 + 3/17 = 37/17

Similarly, 666/306 = 37/17 as above.

777/358: 777 ÷ 21 = 37? 21*37=777? 20*37=740, 1*37=37, total 777 yes. 358 ÷ 21? 21*17=357, close but 358-357=1, not divisible.

358 and 777.

gcd(777,358): 777 ÷ 358 = 2 with remainder 777-716=61

358 ÷ 61 = 5*61=305, remainder 53? 358-305=53

61 ÷ 53 = 1 rem 8

53 ÷ 8 = 6 rem 5

8 ÷ 5 = 1 rem 3

5 ÷ 3 = 1 rem 2

3 ÷ 2 = 1 rem 1

2 ÷ 1 = 2 rem 0 → gcd=1, so fraction is 777/358.

But earlier for others we have 37/17.

37/17 ≈ 2.17647

777/358 ≈ 2.17039 — slightly different.

1147/527 = as above, 1147 ÷ 527.

527 * 2.176 = ? Earlier we had 1147 / 527 = (527*2 + 93)/527 = 2 + 93/527

93/527 = 93÷31 / 527÷31 = 3/17, since 31*17=527, 31*3=93, so 3/17, so 2 + 3/17 = 37/17.

Similarly, 666/306 = 666÷18=37, 306÷18=17, so 37/17.

Now 777/358: let's compute decimal: 777 ÷ 358.

358 * 2.176 = 358*2 = 716, 358*0.176 = let's see, 358*0.1=35.8, 358*0.07=25.06, 358*0.006=2.148, total 35.8+25.06=60.86, +2.148=63.008, so total 716+63.008=779.008 >777, too big.

358*2.17 = 358*2=716, 358*0.17=60.86, total 776.86, and 777 - 776.86=0.14, so approximately 2.1704.

While 37/17 = 2.17647...

So 777/358 ≈ 2.1704, 37/17≈2.1765, not equal.

But for similarity, all three ratios must be equal.

Here, 666/306 = 37/17 ≈2.1765

1147/527 = 37/17 ≈2.1765

But 777/358 ≈2.1704 ≠ 2.1765

So not proportional.

Therefore, not similar by SSS.

Is there any other way? No angles given, so only SSS possible, and it fails.

So not similar.

✔ Answer: C) not similar

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Problem 6)
Two triangles: △ELF and △KLJ? Points: E,L,F and K,L,J, sharing point L.

Angles: at E and at K are marked with red arcs → so ∠E ≅ ∠K

Also, at F and J are marked with red arcs → so ∠F ≅ ∠J

Therefore, two pairs of congruent angles → AA similarity.

Note: the triangles are △EFL and △KJL or something, but vertices: likely △EFL and △KJL, with L common? But in diagram, it's crossed.

Actually, points: E-L-K on a line? And F-L-J on another line? So triangles are △EFL and △KJL, intersecting at L.

Angles at E and K are equal, angles at F and J are equal, so yes, AA.

Even though they are oriented differently, the angles correspond.

So ✔ Answer: D) similar; AA similarity

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Now compiling answers:

1) C
2) B
3) A
4) C
5) C
6) D

Final Answer:
1) C
2) B
3) A
4) C
5) C
6) D