Triangle proofs: fill in the missing step or reason
Grade 10 Math Worksheet

Circle the best answer.
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Given: triangle CDF. Prove: m∠CDF + m∠DCF + m∠CFD = 180. Ladder: 1. GH ∥ DF, through C -- Parallel postulate; 2. ∠GCD ≅ ∠CDF and ∠FCH ≅ ∠CFD -- Alternate interior angles theorem; 3. m∠GCD = m∠CDF and m∠FCH = m∠CFD -- Definition of congruent angles; 4. m∠GCD + m∠DCF + m∠FCH = 180 -- Angle addition postulate and definition of a straight angle; 5. m∠CDF + m∠DCF + m∠CFD = 180 -- ?.
Jun proves the triangle angle sum theorem on a sail CDF. Which reason fills the blank?
- Substitution property
- Angle addition postulate
- Triangle angle sum theorem
- Transitive property
Use this
Given: triangle CDF, with DF extended to G. Prove: m∠CDF + m∠DCF = m∠CFG. Ladder: 1. triangle CDF, with DF extended to G -- Given; 2. m∠CDF + m∠DCF + m∠CFD = 180 -- Triangle angle sum theorem; 3. ∠CFD and ∠CFG are supplementary -- Linear pair postulate; 4. m∠CFD + m∠CFG = 180 -- Definition of supplementary angles; 5. m∠CDF + m∠DCF + m∠CFD = m∠CFD + m∠CFG -- Substitution property; 6. m∠CDF + m∠DCF = m∠CFG -- ?.
Noah proves the exterior angle theorem on a ramp side CDF, with the floor DF running on to G. Which reason fills the blank?
Sketch, not to scale: triangle CDF with C at the top, D bottom left and F bottom right; side DF is extended past F to G. Marked: the outside angle ∠CFG shaded. Given: △CDF, with DF extended to G. Prove: m∠CDF + m∠DCF = m∠CFG. Ladder: 1. △CDF, with DF extended to G — Given; 2. m∠CDF + m∠DCF + m∠CFD = 180 — Triangle angle sum theorem; 3. ∠CFD and ∠CFG are supplementary — Linear pair postulate; 4. m∠CFD + m∠CFG = 180 — Definition of supplementary angles; 5. m∠CDF + m∠DCF + m∠CFD = m∠CFD + m∠CFG — Substitution property; 6. m∠CDF + m∠DCF = m∠CFG — ?. Sketch, not to scale: triangle CDF with C at the top, D bottom left and F bottom right; side DF is extended past F to G. Marked: the outside angle ∠CFG shaded. Given: △CDF, with DF extended to G. Prove: m∠CDF + m∠DCF = m∠CFG. Ladder: 1. △CDF, with DF extended to G — Given; 2. m∠CDF + m∠DCF + m∠CFD = 180 — Triangle angle sum theorem; 3. ∠CFD and ∠CFG are supplementary — Linear pair postulate; 4. m∠CFD + m∠CFG = 180 — Definition of supplementary angles; 5. m∠CDF + m∠DCF + m∠CFD = m∠CFD + m∠CFG — Substitution property; 6. m∠CDF + m∠DCF = m∠CFG — ?. ∠CFG C D F G Not to scale. - Subtraction property of equality
- Exterior angle theorem
- Given
- Addition property of equality
Use this
Given: triangle SUV, with UV extended to W. Prove: m∠SUV + m∠USV = m∠SVW. Ladder: 1. triangle SUV, with UV extended to W -- Given; 2. m∠SUV + m∠USV + m∠SVU = 180 -- Triangle angle sum theorem; 3. ∠SVU and ∠SVW are supplementary -- Linear pair postulate; 4. m∠SVU + m∠SVW = 180 -- Definition of supplementary angles; 5. m∠SUV + m∠USV + m∠SVU = m∠SVU + m∠SVW -- Substitution property; 6. m∠SUV + m∠USV = m∠SVW -- ?.
Pia proves the exterior angle theorem on a road sign SUV, extending the bottom edge UV to W. Which reason fills the blank?
Sketch, not to scale: triangle SUV with S at the top, U bottom left and V bottom right; side UV is extended past V to W. Marked: the outside angle ∠SVW shaded. Given: △SUV, with UV extended to W. Prove: m∠SUV + m∠USV = m∠SVW. Ladder: 1. △SUV, with UV extended to W — Given; 2. m∠SUV + m∠USV + m∠SVU = 180 — Triangle angle sum theorem; 3. ∠SVU and ∠SVW are supplementary — Linear pair postulate; 4. m∠SVU + m∠SVW = 180 — Definition of supplementary angles; 5. m∠SUV + m∠USV + m∠SVU = m∠SVU + m∠SVW — Substitution property; 6. m∠SUV + m∠USV = m∠SVW — ?. Sketch, not to scale: triangle SUV with S at the top, U bottom left and V bottom right; side UV is extended past V to W. Marked: the outside angle ∠SVW shaded. Given: △SUV, with UV extended to W. Prove: m∠SUV + m∠USV = m∠SVW. Ladder: 1. △SUV, with UV extended to W — Given; 2. m∠SUV + m∠USV + m∠SVU = 180 — Triangle angle sum theorem; 3. ∠SVU and ∠SVW are supplementary — Linear pair postulate; 4. m∠SVU + m∠SVW = 180 — Definition of supplementary angles; 5. m∠SUV + m∠USV + m∠SVU = m∠SVU + m∠SVW — Substitution property; 6. m∠SUV + m∠USV = m∠SVW — ?. ∠SVW S U V W Not to scale. - Subtraction property of equality
- Exterior angle theorem
- Addition property of equality
- Given
Use this
Given: SU ≅ SW, and Y is the midpoint of UW. Prove: ∠SUY ≅ ∠SWY. Ladder: 1. SU ≅ SW -- Given; 2. Y is the midpoint of UW -- Given; 3. UY ≅ WY -- Definition of midpoint; 4. SY ≅ SY -- Reflexive property; 5. triangle SUY ≅ triangle SWY -- SSS; 6. ∠SUY ≅ ∠SWY -- ?.
Ari proves the base angles theorem on a roof SUW, with a beam from S to the middle Y of UW. Which reason fills the blank?
Sketch, not to scale: triangle SUW with S at the top and base UW; Y is on UW, and segment SY is drawn. Marked: SU and SW with one tick each; UY and YW with two ticks each. Given: SU ≅ SW, and Y is the midpoint of UW. Prove: ∠SUY ≅ ∠SWY. Ladder: 1. SU ≅ SW — Given; 2. Y is the midpoint of UW — Given; 3. UY ≅ WY — Definition of midpoint; 4. SY ≅ SY — Reflexive property; 5. △SUY ≅ △SWY — SSS; 6. ∠SUY ≅ ∠SWY — ?. Sketch, not to scale: triangle SUW with S at the top and base UW; Y is on UW, and segment SY is drawn. Marked: SU and SW with one tick each; UY and YW with two ticks each. Given: SU ≅ SW, and Y is the midpoint of UW. Prove: ∠SUY ≅ ∠SWY. Ladder: 1. SU ≅ SW — Given; 2. Y is the midpoint of UW — Given; 3. UY ≅ WY — Definition of midpoint; 4. SY ≅ SY — Reflexive property; 5. △SUY ≅ △SWY — SSS; 6. ∠SUY ≅ ∠SWY — ?. S U Y W Not to scale. - Base angles theorem
- Given
- CPCTC
- SSS
Use this
Given: triangle VDH. Prove: m∠VDH + m∠DVH + m∠VHD = 180. Ladder: 1. CF ∥ DH, through V -- Parallel postulate; 2. ? -- Alternate interior angles theorem; 3. m∠CVD = m∠VDH and m∠HVF = m∠VHD -- Definition of congruent angles; 4. m∠CVD + m∠DVH + m∠HVF = 180 -- Angle addition postulate and definition of a straight angle; 5. m∠VDH + m∠DVH + m∠VHD = 180 -- Substitution property.
Luz proves the triangle angle sum theorem on a roof VDH. Which statement fills the blank?
Sketch, not to scale: triangle VDH with V at the top, D bottom left and H bottom right; line CF is drawn through V (C on the left, F on the right). Marked: parallel arrows on CF and DH. Given: △VDH. Prove: m∠VDH + m∠DVH + m∠VHD = 180. Ladder: 1. CF ∥ DH, through V — Parallel postulate; 2. ? — Alternate interior angles theorem; 3. m∠CVD = m∠VDH and m∠HVF = m∠VHD — Definition of congruent angles; 4. m∠CVD + m∠DVH + m∠HVF = 180 — Angle addition postulate and definition of a straight angle; 5. m∠VDH + m∠DVH + m∠VHD = 180 — Substitution property. Sketch, not to scale: triangle VDH with V at the top, D bottom left and H bottom right; line CF is drawn through V (C on the left, F on the right). Marked: parallel arrows on CF and DH. Given: △VDH. Prove: m∠VDH + m∠DVH + m∠VHD = 180. Ladder: 1. CF ∥ DH, through V — Parallel postulate; 2. ? — Alternate interior angles theorem; 3. m∠CVD = m∠VDH and m∠HVF = m∠VHD — Definition of congruent angles; 4. m∠CVD + m∠DVH + m∠HVF = 180 — Angle addition postulate and definition of a straight angle; 5. m∠VDH + m∠DVH + m∠VHD = 180 — Substitution property. V D H C F Not to scale. - m∠VDH + m∠DVH + m∠VHD = 180
- ∠CVD ≅ ∠VHD and ∠HVF ≅ ∠VDH
- ∠CVD ≅ ∠DVH and ∠HVF ≅ ∠DVH
- ∠CVD ≅ ∠VDH and ∠HVF ≅ ∠VHD
Sketch, not to scale
Triangle GSY with G at the top, S bottom left and Y bottom right; line UW is drawn through G (U on the left, W on the right). Marked: parallel arrows on UW and SY. Given: △GSY. Prove: m∠GSY + m∠SGY + m∠GYS = 180. Ladder: 1. UW ∥ SY, through G — Parallel postulate; 2. ? — Alternate interior angles theorem; 3. m∠UGS = m∠GSY and m∠YGW = m∠GYS — Definition of congruent angles; 4. m∠UGS + m∠SGY + m∠YGW = 180 — Angle addition postulate and definition of a straight angle; 5. m∠GSY + m∠SGY + m∠GYS = 180 — Substitution property.
Milo proves the triangle angle sum theorem on a tent front GSY. Which statement fills the blank?
- m∠GSY + m∠SGY + m∠GYS = 180
- ∠UGS ≅ ∠GYS and ∠YGW ≅ ∠GSY
- ∠UGS ≅ ∠SGY and ∠YGW ≅ ∠SGY
- ∠UGS ≅ ∠GSY and ∠YGW ≅ ∠GYS
Use this
Given: triangle HQY, with QY extended to D. Prove: m∠HQY + m∠QHY = m∠HYD. Ladder: 1. triangle HQY, with QY extended to D -- Given; 2. m∠HQY + m∠QHY + m∠HYQ = 180 -- Triangle angle sum theorem; 3. ? -- Linear pair postulate; 4. m∠HYQ + m∠HYD = 180 -- Definition of supplementary angles; 5. m∠HQY + m∠QHY + m∠HYQ = m∠HYQ + m∠HYD -- Substitution property; 6. m∠HQY + m∠QHY = m∠HYD -- Subtraction property of equality.
Ren proves the exterior angle theorem on a ramp side HQY, with the floor running on to D. Which statement fills the blank?
Sketch, not to scale: triangle HQY with H at the top, Q bottom left and Y bottom right; side QY is extended past Y to D. Marked: the outside angle ∠HYD shaded. Given: △HQY, with QY extended to D. Prove: m∠HQY + m∠QHY = m∠HYD. Ladder: 1. △HQY, with QY extended to D — Given; 2. m∠HQY + m∠QHY + m∠HYQ = 180 — Triangle angle sum theorem; 3. ? — Linear pair postulate; 4. m∠HYQ + m∠HYD = 180 — Definition of supplementary angles; 5. m∠HQY + m∠QHY + m∠HYQ = m∠HYQ + m∠HYD — Substitution property; 6. m∠HQY + m∠QHY = m∠HYD — Subtraction property of equality. Sketch, not to scale: triangle HQY with H at the top, Q bottom left and Y bottom right; side QY is extended past Y to D. Marked: the outside angle ∠HYD shaded. Given: △HQY, with QY extended to D. Prove: m∠HQY + m∠QHY = m∠HYD. Ladder: 1. △HQY, with QY extended to D — Given; 2. m∠HQY + m∠QHY + m∠HYQ = 180 — Triangle angle sum theorem; 3. ? — Linear pair postulate; 4. m∠HYQ + m∠HYD = 180 — Definition of supplementary angles; 5. m∠HQY + m∠QHY + m∠HYQ = m∠HYQ + m∠HYD — Substitution property; 6. m∠HQY + m∠QHY = m∠HYD — Subtraction property of equality. ∠HYD H Q Y D Not to scale. - ∠HYQ and ∠HYD are complementary
- m∠HQY + m∠QHY = m∠HYD
- ∠HYQ and ∠HYD are supplementary
- ∠HYQ ≅ ∠HYD
Use this
Sketch, not to scale: triangle GCW with G at the top, C bottom left and W bottom right; side CW is extended past W to F. Marked: the outside angle ∠GWF shaded. Given: triangle GCW, with CW extended to F. Prove: m∠GCW + m∠CGW = m∠GWF. Ladder: 1. triangle GCW, with CW extended to F -- Given; 2. m∠GCW + m∠CGW + m∠GWC = 180 -- Triangle angle sum theorem; 3. ? -- Linear pair postulate; 4. m∠GWC + m∠GWF = 180 -- Definition of supplementary angles; 5. m∠GCW + m∠CGW + m∠GWC = m∠GWC + m∠GWF -- Substitution property; 6. m∠GCW + m∠CGW = m∠GWF -- Subtraction property of equality.
Tala proves the exterior angle theorem on a sandwich half GCW, with edge CW drawn on to F. Which statement fills the blank?
Sketch, not to scale: triangle GCW with G at the top, C bottom left and W bottom right; side CW is extended past W to F. Marked: the outside angle ∠GWF shaded. Given: △GCW, with CW extended to F. Prove: m∠GCW + m∠CGW = m∠GWF. Ladder: 1. △GCW, with CW extended to F — Given; 2. m∠GCW + m∠CGW + m∠GWC = 180 — Triangle angle sum theorem; 3. ? — Linear pair postulate; 4. m∠GWC + m∠GWF = 180 — Definition of supplementary angles; 5. m∠GCW + m∠CGW + m∠GWC = m∠GWC + m∠GWF — Substitution property; 6. m∠GCW + m∠CGW = m∠GWF — Subtraction property of equality. Sketch, not to scale: triangle GCW with G at the top, C bottom left and W bottom right; side CW is extended past W to F. Marked: the outside angle ∠GWF shaded. Given: △GCW, with CW extended to F. Prove: m∠GCW + m∠CGW = m∠GWF. Ladder: 1. △GCW, with CW extended to F — Given; 2. m∠GCW + m∠CGW + m∠GWC = 180 — Triangle angle sum theorem; 3. ? — Linear pair postulate; 4. m∠GWC + m∠GWF = 180 — Definition of supplementary angles; 5. m∠GCW + m∠CGW + m∠GWC = m∠GWC + m∠GWF — Substitution property; 6. m∠GCW + m∠CGW = m∠GWF — Subtraction property of equality. ∠GWF ∠GWF G C W F Not to scale. - ∠GWC and ∠GWF are supplementary
- ∠GWC ≅ ∠GWF
- ∠GWC and ∠GWF are complementary
- m∠GCW + m∠CGW = m∠GWF
Bo proves the base angles theorem on a road sign CHV, marking the middle Q of HV. Which statement fills the blank?
Sketch, not to scale: triangle CHV with C at the top and base HV; Q is on HV, and segment CQ is drawn. Marked: CH and CV with one tick each; HQ and QV with two ticks each. Given: CH ≅ CV, and Q is the midpoint of HV. Prove: ∠CHQ ≅ ∠CVQ. Ladder: 1. CH ≅ CV — Given; 2. Q is the midpoint of HV — Given; 3. HQ ≅ VQ — Definition of midpoint; 4. CQ ≅ CQ — Reflexive property; 5. ? — SSS; 6. ∠CHQ ≅ ∠CVQ — CPCTC. Sketch, not to scale: triangle CHV with C at the top and base HV; Q is on HV, and segment CQ is drawn. Marked: CH and CV with one tick each; HQ and QV with two ticks each. Given: CH ≅ CV, and Q is the midpoint of HV. Prove: ∠CHQ ≅ ∠CVQ. Ladder: 1. CH ≅ CV — Given; 2. Q is the midpoint of HV — Given; 3. HQ ≅ VQ — Definition of midpoint; 4. CQ ≅ CQ — Reflexive property; 5. ? — SSS; 6. ∠CHQ ≅ ∠CVQ — CPCTC. H V C Q Not to scale - △CHV ≅ △CVQ
- ∠CHQ ≅ ∠CVQ
- △CHQ ≅ △CVQ
- △CHQ ≅ △QVC
Esme proves the base angles theorem on a pyramid face FDY, marking the middle G of DY. Which statement fills the blank?
Sketch, not to scale: triangle FDY with F at the top and base DY; G is on DY, and segment FG is drawn. Marked: FD and FY with one tick each; DG and GY with two ticks each. Given: FD ≅ FY, and G is the midpoint of DY. Prove: ∠FDG ≅ ∠FYG. Ladder: 1. FD ≅ FY — Given; 2. G is the midpoint of DY — Given; 3. DG ≅ YG — Definition of midpoint; 4. FG ≅ FG — Reflexive property; 5. ? — SSS; 6. ∠FDG ≅ ∠FYG — CPCTC. Sketch, not to scale: triangle FDY with F at the top and base DY; G is on DY, and segment FG is drawn. Marked: FD and FY with one tick each; DG and GY with two ticks each. Given: FD ≅ FY, and G is the midpoint of DY. Prove: ∠FDG ≅ ∠FYG. Ladder: 1. FD ≅ FY — Given; 2. G is the midpoint of DY — Given; 3. DG ≅ YG — Definition of midpoint; 4. FG ≅ FG — Reflexive property; 5. ? — SSS; 6. ∠FDG ≅ ∠FYG — CPCTC. D Y F G Not to scale - △FDY ≅ △FYG
- △FDG ≅ △GYF
- ∠FDG ≅ ∠FYG
- △FDG ≅ △FYG