Triangle Congruence Worksheet - Fill Online, Printable, Fillable ... - Free Printable

Let’s go through each triangle pair one by one. We’re looking for the postulate or theorem that proves the triangles are congruent — meaning they have the same size and shape.

We’ll use these common rules:
- SSS (Side-Side-Side): All three sides match.
- SAS (Side-Angle-Side): Two sides and the angle between them match.
- ASA (Angle-Side-Angle): Two angles and the side between them match.
- AAS (Angle-Angle-Side): Two angles and a non-included side match.
- HL (Hypotenuse-Leg): Only for right triangles — hypotenuse and one leg match.

Now let’s solve each problem:

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1. Triangles GFI and GHI
→ GI is shared (common side)
→ ∠GIF = ∠GIH (both marked as right angles? Wait — actually, in diagram 1, it looks like GI is perpendicular to FH, so both angles at I are right angles → 90°)
→ Also, FI ≅ HI (marked with tick marks)
So we have: Side (FI=HI), Angle (∠F IG = ∠HIG = 90°), Side (GI=GI) → SAS

Wait — actually, looking again: The two triangles share side GI. FI and HI are marked equal. And the angles at I are both right angles (since there’s a square symbol). So yes — SAS.

But wait — another way: If you look at triangle GFI and GHI, you have:
- GI = GI (reflexive)
- FI = HI (given by tick marks)
- ∠GIF = ∠GIH = 90° (right angle symbols)

That’s SAS — because the angle is *between* the two sides.

✔ Answer: SAS

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2. Quadrilateral DEFG — diagonals intersect? Actually, it shows triangles DEF and DGF? Or maybe DEG and FEG? Let me see…

Actually, it’s quadrilateral DEFG with diagonal DF. So triangles DEF and DGF? No — probably triangles DEG and FEG? Wait — better to label:

Points: D, E, F, G — connected as a kite? Diagonal DF is drawn.

Triangles: △DEF and △DGF? But no — likely △DEG and △FEG? Hmm.

Looking at markings:
- DE ≅ FE (one tick)
- DG ≅ FG (two ticks)
- EG is common

So triangles DEG and FEG:
- DE = FE
- DG = FG
- EG = EG

All three sides match → SSS

✔ Answer: SSS

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3. Two triangles sharing point C: △ABC and △EDC
Markings:
- BC ≅ DC (tick marks)
- AC ≅ EC (tick marks)
- ∠ACB ≅ ∠ECD (vertical angles — always equal!)

So: Side (BC=DC), Angle (∠C), Side (AC=EC) → SAS

✔ Answer: SAS

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4. Square ABCD with diagonal BD
Triangles: △ABD and △CBD
In a square:
- AB = CB (sides of square)
- AD = CD (sides of square)
- BD = BD (common)

Also, all angles are 90°, but we don’t need that.

Three sides match → SSS

Alternatively, since it’s a square, diagonal splits into two congruent right triangles — could also be HL if we consider right angles, but SSS works too.

But note: In square, AB = CB, AD = CD, BD common → SSS.

✔ Answer: SSS

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5. Quadrilateral JKLM with diagonal KM
Triangles: △JKM and △LKM
Markings:
- ∠J ≅ ∠L (angle marks)
- ∠JKM ≅ ∠LKM (angle marks at K)
- KM is common

So: Two angles and included side → ASA

Wait — angles at J and L, and angles at K (on either side of KM), and side KM is between them? Let’s see:

In △JKM and △LKM:
- ∠J = ∠L
- ∠JKM = ∠LKM
- Side KM is common → and it’s between those two angles? Yes — in each triangle, KM is between angle at K and angle at M? Wait — no.

Actually, in △JKM: angles at J and K, side KM is opposite angle J? Not quite.

Better: We have two angles and a non-included side? Let’s list:

△JKM and △LKM:
- ∠J ≅ ∠L
- ∠JKM ≅ ∠LKM
- Side KM is common → but KM is adjacent to ∠K in both, but not between ∠J and K.

Actually, this is AAS: two angles and a non-included side.

Because:
- ∠J = ∠L
- ∠JKM = ∠LKM
- Side KM is common → and it’s not between the two angles in the sense of ASA? Wait — in ASA, the side must be *between* the two angles.

Here, in △JKM, the two angles are at J and K, and side KM is opposite angle J? No — side KM connects K and M.

Perhaps better to think: we have two angles and any side — if two angles are equal, the third is automatically equal, so AAS or ASA both work depending on which side.

But here, side KM is common, and it’s the side *not* between the two given angles? Let’s sketch mentally:

Triangle JKM: vertices J, K, M
Angles given: at J and at K
Side KM is between K and M — so it’s adjacent to angle K, but not between angle J and angle K.

Actually, standard rule: if you have two angles and a side that is *not* between them, it’s AAS.

In this case, side KM is opposite angle J in triangle JKM? No — in triangle JKM, side opposite angle J is KM? Yes! Because angle J is at vertex J, opposite side is KM.

Similarly, in triangle LKM, angle L is at L, opposite side is KM.

So we have:
- ∠J ≅ ∠L
- ∠JKM ≅ ∠LKM
- Side KM ≅ KM (and it’s opposite the first angle in each triangle)

This fits AAS — two angles and a non-included side.

Some curricula accept AAS as valid.

✔ Answer: AAS

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6. Quadrilateral NOPQ with diagonal OQ
Triangles: △NOQ and △POQ
Markings:
- NO ≅ PO (tick marks)
- NQ ≅ PQ (tick marks)
- OQ is common

So all three sides match → SSS

✔ Answer: SSS

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7. Two triangles sharing point C: △ABC and △EDC
Markings:
- AC ≅ EC (tick marks)
- BC ≅ DC (tick marks)
- ∠ACB ≅ ∠ECD (vertical angles)

Same as #3 → SAS

✔ Answer: SAS

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8. Triangle FGH with altitude GI to base FH
So GI ⊥ FH → right angles at I
Also, FI ≅ HI (tick marks)
GI is common

So triangles FIG and HIG:
- FI = HI
- GI = GI
- ∠FIG = ∠HIG = 90°

→ SAS (side-angle-side)

Could also be HL since they are right triangles: hypotenuse FG and HG? But we don’t know if FG=HG yet — that’s what we’re trying to prove.

But with FI=HI, GI=GI, and right angles → SAS is direct.

✔ Answer: SAS

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9. Quadrilateral KILM with diagonal IM
Triangles: △KIM and △LIM
Markings:
- KI ≅ LI (tick marks)
- KM ≅ LM (tick marks)
- IM is common

→ All three sides match → SSS

✔ Answer: SSS

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10. Quadrilateral MNOP with diagonal NP
Triangles: △MNP and △ONP
Markings:
- MN ≅ ON (tick marks)
- MP ≅ OP (tick marks)
- NP is common

→ SSS

✔ Answer: SSS

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11. Quadrilateral JKLM with diagonal KM
Triangles: △JKM and △LKM
Markings:
- ∠J ≅ ∠L (angle marks)
- ∠JMK ≅ ∠LMK (angle marks at M)
- KM is common

Same as #5 → two angles and included side? Let’s see:

In △JKM and △LKM:
- ∠J = ∠L
- ∠JMK = ∠LMK
- Side KM is common → and it’s between the two angles? In each triangle, KM is between angle at M and angle at K? Wait — angle at M is ∠JMK, which is at M, and angle at K is... we don’t have angle at K marked.

Actually, we have angles at J and at M, and side KM.

In △JKM: angles at J and M, side KM is between them? Yes — because side KM connects K and M, so in triangle JKM, side KM is between angle at K and angle at M? No — angle at M is ∠JMK, which is formed by sides JM and KM.

Actually, side KM is adjacent to angle at M, but not between angle J and angle M.

Standard: if you have two angles and the side between them → ASA.

Here, in △JKM, the two angles are at J and M. The side between them would be JM — but we don’t have that marked.

We have side KM, which is not between angles J and M — it’s opposite angle J? In triangle JKM, side opposite angle J is KM — yes.

So again, two angles and a non-included side → AAS

Same as #5.

✔ Answer: AAS

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12. Rectangle RSTU with diagonal RT
Triangles: △RST and △UTR? Or △RSU and △TSU? Better: diagonal RT divides rectangle into △RST and △RUT? Points: R,S,T,U — probably R-S-T-U clockwise.

Diagonal from R to T.

Triangles: △RST and △RUT? No — should be △RST and △URT? Standard: △RST and △RUT don't make sense.

Actually, diagonal RT creates △RST and △RUT? Let's define:

Assume rectangle R-S-T-U, so sides RS, ST, TU, UR.

Diagonal RT connects R to T.

Then triangles are △RST and △RUT? No — △RST has points R,S,T — that’s half the rectangle.

Other triangle is △RUT? Points R,U,T — yes.

So △RST and △RUT.

In rectangle:
- RS = UT (opposite sides)
- ST = RU (opposite sides)
- RT = RT (common)

Also, all angles 90°.

So three sides match → SSS

Could also be HL: both right triangles, hypotenuse RT common, and legs RS=UT, etc.

But SSS is straightforward.

✔ Answer: SSS

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Now let’s compile all answers:

1. SAS
2. SSS
3. SAS
4. SSS
5. AAS
6. SSS
7. SAS
8. SAS
9. SSS
10. SSS
11. AAS
12. SSS

Final Answer:
1. SAS
2. SSS
3. SAS
4. SSS
5. AAS
6. SSS
7. SAS
8. SAS
9. SSS
10. SSS
11. AAS
12. SSS