Use congruent triangles to find missing sides and angles
Grade 10 Math Worksheet

Circle the best answer.
Ari proves CB = CD for the card. Which lines complete the proof?
- AC = AC (reflexive); ASA gives △ABC ≅ △ADC; CB = CD by CPCTC
- CB and CD look equal; AC = AC; SSS gives △ABC ≅ △ADC
- AC = AC (reflexive); SAS gives △ABC ≅ △ADC; CB = CD by CPCTC
- ∠BAC = ∠DAC, so CB = CD by CPCTC
Bo proves AC = BD. Which lines complete the proof?
Segments AB and CD cross at M. M is the midpoint of AB and of CD. Segments AB and CD cross at M. M is the midpoint of AB and of CD. A B C D M - AM = BM, CM = DM, ∠AMC = ∠BMD (vertical); ASA gives △AMC ≅ △BMD; AC = BD by CPCTC
- ∠AMC = ∠BMD (vertical), so AC = BD by CPCTC
- AM = BM, CM = DM, ∠AMC = ∠BMD (vertical); SAS gives △AMC ≅ △BMD; AC = BD by CPCTC
- AC = BD is assumed, so △AMC ≅ △BMD by SSS
Jun proves AB = DC. Which lines complete the proof?
Parallelogram ABCD with diagonal AC, so AB ∥ DC and AD ∥ BC. Parallelogram ABCD with diagonal AC, so AB ∥ DC and AD ∥ BC. Parallelogram ABCD: AB ∥ DC and AD ∥ BC. A B C D - Opposite sides of a parallelogram are equal, so AB = DC at once
- Alternate angles and AC = AC give ASA, so △ABC ≅ △CDA and AB = DC
- △ABC ≅ △ACD by ASA, so AB = AC by CPCTC
- Alternate angles and AC = AC give SAS, so △ABC ≅ △CDA and AB = DC
Read this first
Use this
Proof: 1. AB = AC (given). 2. ∠BAD = ∠CAD (AD bisects ∠BAC). 3. ? 4. triangle ABD ≅ triangle ACD (SAS). 5. BD = CD (CPCTC), so D is the midpoint.
Kira proves D is the midpoint of BC. Which line is missing from the proof?
Triangle ABC with D on BC. Proof: 1. AB = AC (given). 2. ∠BAD = ∠CAD (AD bisects ∠BAC). 3. ? 4. △ABD ≅ △ACD (SAS). 5. BD = CD (CPCTC), so D is the midpoint. Triangle ABC with D on BC. Proof: 1. AB = AC (given). 2. ∠BAD = ∠CAD (AD bisects ∠BAC). 3. ? 4. △ABD ≅ △ACD (SAS). 5. BD = CD (CPCTC), so D is the midpoint. A B C D - ∠ADB = ∠ADC = 90°, since AD looks perpendicular
- BD = CD (definition of midpoint)
- AD = AD (reflexive property)
- ∠ABD = ∠ACD, pairing the wrong vertices
For A(1, 1), B(5, 2), C(6, 6) and D(2, 5), Luz proves ABCD is a parallelogram. Which computed fact is enough on its own?
Quadrilateral ABCD with A(1, 1), B(5, 2), C(6, 6) and D(2, 5) on a grid. Slope of AB = 1/4 and of DC = 1/4; AB and DC both have length √17. Quadrilateral ABCD with A(1, 1), B(5, 2), C(6, 6) and D(2, 5) on a grid. Slope of AB = 1/4 and of DC = 1/4; AB and DC both have length √17. 1 2 3 4 5 6 7 1 2 3 4 5 6 7 0 A B C D Slope of AB = 1/4 and of DC = 1/4; AB and DC both have length √17. - AB = DC = √17 and nothing else
- AB ∥ DC: both slopes are 1/4 and nothing else
- AB ∥ DC and AB = DC: both slopes are 1/4 and both lengths are √17
- Not a parallelogram, because the diagonals are unequal
△ABC has A(1, 1), B(4, 1), C(1, 3). △A′B′C′ has A′(−1, 1), B′(−1, 4), C′(−3, 1). Milo says they are congruent. Which reasoning proves it?
△ABC with A(1, 1), B(4, 1), C(1, 3) and △A′B′C′ with A′(−1, 1), B′(−1, 4), C′(−3, 1) on a grid. △ABC with A(1, 1), B(4, 1), C(1, 3) and △A′B′C′ with A′(−1, 1), B′(−1, 4), C′(−3, 1) on a grid. −4 −3 −2 −1 1 2 3 4 5 1 2 3 4 5 0 A B C A' B' C' - Their angles match, so AAA makes them congruent
- Both are right triangles, so they are congruent
- They look the same size on the grid
- A 90° turn about the origin maps A, B, C to A′, B′, C′; turns keep lengths and angles
△ABC has A(0, 0), B(4, 0), C(0, 3). △A′B′C′ has A′(0, 0), B′(8, 0), C′(0, 6). Noah compares them. Which statement is justified?
△ABC with A(0, 0), B(4, 0), C(0, 3) and △A′B′C′ with A′(0, 0), B′(8, 0), C′(0, 6) on a grid. △ABC with A(0, 0), B(4, 0), C(0, 3) and △A′B′C′ with A′(0, 0), B′(8, 0), C′(0, 6) on a grid. −1 1 2 3 4 5 6 7 8 9 −1 1 2 3 4 5 6 7 0 A B C A' B' C' - Similar when a rigid motion maps one onto the other
- A dilation from the origin with scale factor 2 maps one to the other: similar, not congruent
- Not similar: the side lengths differ
- Congruent: both are right triangles at the origin
Two triangles with two equal sides and one equal angle are always congruent. Pia looks for numbers that break the claim. Which pair of triangles does it?
- Right triangles with hypotenuse 13 and one leg 5: two triangles fit
- AB = 7, BC = 5, ∠A = 30° (opposite BC): AC is about 9.6 or 2.5
- AB = 5, BC = 7 and ∠A = 30° (opposite the longer side): two triangles fit
- AB = 7, AC = 5 and ∠A = 30° (between the sides): the triangle is not fixed
Two triangles with equal areas are congruent. Ren looks for a pair that breaks the claim. Which pair does it?
- Right triangles with hypotenuse 10 and area 24 in both
- Equilateral triangles with sides 4 and 4.1: the areas are equal
- Right triangles with legs 6 and 4, and legs 4 and 6: both have area 12
- Right triangles with legs 6 and 4, and legs 8 and 3: both have area 12
Two triangles with all three angles equal are congruent. Zee looks for a pair that breaks the claim. Which pair does it?
- Sides 5, 5, 6 and sides 5, 5, 8: two sides match in each
- Sides 3, 4, 5 and sides 4, 3, 5: same angles, same size
- Sides 3, 4, 5 and sides 6, 8, 10: same angles, different sizes
- Sides 3, 4, 5 and sides 4, 5, 6: both look nearly right-angled

